Сканави (А) 2.1-2.15

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№ 2.1 Өрнекті ықшамдаңыз: $\dfrac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }}:\dfrac{1}{{{x^2} - \sqrt x }}$

Шешуі: ММЖ: $0 \lt x \ne 1$ $${\dfrac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }}:\dfrac{1}{{{x^2} - \sqrt x }} = \dfrac{{\sqrt x + 1}}{{\sqrt x (x + \sqrt x + 1)}} \cdot \dfrac{{\sqrt x (x\sqrt x - 1)}}{1} = }$$ $${ = \dfrac{{(\sqrt x + 1)(\sqrt x - 1)}}{{\sqrt x (x + \sqrt x + 1)(\sqrt x - 1)}} \cdot \dfrac{{\sqrt x (x\sqrt x - 1)}}{1} = \dfrac{{x - 1}}{{\sqrt x (x\sqrt x - 1)}} \times }$$ $${ \times \dfrac{{\sqrt x (x\sqrt x - 1)}}{1} = x - 1}$$

№ 2.2 Өрнекті ықшамдаңыз: $\left( {{{(\sqrt[4]{p} - \sqrt[4]{q})}^{ - 2}} + {{(\sqrt[4]{p} + \sqrt[4]{q})}^{ - 2}}} \right):\dfrac{{\sqrt p + \sqrt q }}{{p - q}}$

Шешуі: ММЖ: $p \ne q$ $${ = \left( {\dfrac{1}{{{{(\sqrt[4]{p} - \sqrt[4]{q})}^2}}} + \dfrac{1}{{{{(\sqrt[4]{p} + \sqrt[4]{q})}^2}}}} \right) \times \dfrac{{p - q}}{{\sqrt p + \sqrt q }} = }$$ $${ = \dfrac{{{{(\sqrt[4]{p} + \sqrt[4]{q})}^2} + {{(\sqrt[4]{p} - \sqrt[4]{q})}^2}}}{{{{(\sqrt p - \sqrt q )}^2}}} \cdot \dfrac{{(\sqrt p - \sqrt q )(\sqrt p + \sqrt q )}}{{\sqrt p + \sqrt q }} = }$$ $${ = \dfrac{{\sqrt p + 2\sqrt[4]{{pq}} + \sqrt q + \sqrt p - 2\sqrt[4]{{pq}} + \sqrt q }}{{\sqrt p - \sqrt q }} = \dfrac{{2(\sqrt p + \sqrt q )}}{{\sqrt p - \sqrt q }} = }$$ $${ = \dfrac{{2(\sqrt p + \sqrt q )(\sqrt p + \sqrt q )}}{{(\sqrt p - \sqrt q )(\sqrt p + \sqrt q )}} = \dfrac{{2{{(\sqrt p + \sqrt q )}^2}}}{{p - q}}}$$

№ 2.3Өрнекті ықшамдаңыз: $\dfrac{{\left( {\sqrt {{a^2} + a\sqrt {{a^2} - {b^2}} } } \right) - {{\left( {\sqrt {{a^2} - a\sqrt {{a^2} - {b^2}} } } \right)}^2}}}{{2\sqrt {{a^3}b} }} \cdot \left( {\sqrt {\dfrac{a}{b}} + \sqrt {\dfrac{b}{a}} - 2} \right),\,\,(a \gt b \gt 0)$

Шешуі: $${X = \dfrac{{\left( {\sqrt {\left. {{a^2} + a\sqrt {{a^2} - {b^2}} } \right)} - {{\left( {\sqrt {{a^2} - a\sqrt {{a^2} - {b^2}} } } \right)}^2}} \right.}}{{2\sqrt {{a^3}b} }} = }$$ $${ = \dfrac{{{{\left( {\sqrt {a\left( {a + \sqrt {{a^2} - {b^2}} } \right)} - \sqrt {a\left( {a - \sqrt {{a^2} - {b^2}} } \right)} } \right)}^2}}}{{2a\sqrt {ab} }} = }$$ $${ = \dfrac{{{{\left( {\sqrt a \left( {\sqrt {a + \sqrt {{a^2} - {b^2}} } - \sqrt {a - \sqrt {{a^2} - {b^2}} } } \right)} \right)}^2}}}{{2a\sqrt {ab} }} = }$$ $${ = \dfrac{{a\left( {a + \sqrt {{a^2} - {b^2}} - 2\sqrt {{a^2} - {a^2} + {b^2}} + a - \sqrt {{a^2} - {b^2}} } \right)}}{{2a\sqrt {ab} }} = \dfrac{{2a - 2b}}{{2\sqrt {ab} }} = \dfrac{{a - b}}{{\sqrt {ab} }}}$$ $${Y = \sqrt {\dfrac{a}{b}} + \sqrt {\dfrac{b}{a}} - 2 = \dfrac{{\sqrt a }}{{\sqrt b }} + \dfrac{{\sqrt b }}{{\sqrt a }} - 2 = \dfrac{{{{(\sqrt a )}^2} - 2\sqrt {ab} + {{(\sqrt b )}^2}}}{{\sqrt {ab} }} = \dfrac{{{{(\sqrt a - \sqrt b )}^2}}}{{\sqrt {ab} }}}$$ онда $${X:Y = \dfrac{{a - b}}{{\sqrt {ab} }}:\dfrac{{{{(\sqrt a - \sqrt b )}^2}}}{{\sqrt {ab} }} = \dfrac{{(a - b) \cdot \sqrt {ab} }}{{\sqrt {ab} \cdot {{(\sqrt a - \sqrt b )}^2}}} = \dfrac{{a - b}}{{{{(\sqrt a - \sqrt b )}^2}}} = }$$ $${ = \dfrac{{(a - b){{(\sqrt a + \sqrt b )}^2}}}{{{{(\sqrt a - \sqrt b )}^2}{{(\sqrt a + \sqrt b )}^2}}} = \dfrac{{(a - b){{(\sqrt a + \sqrt b )}^2}}}{{{{((\sqrt a - \sqrt b )(\sqrt a + \sqrt b ))}^2}}} = \dfrac{{(a - b){{(\sqrt a + \sqrt b )}^2}}}{{{{(a - b)}^2}}} = }$$ $${ = \dfrac{{{{(\sqrt a + \sqrt b )}^2}}}{{a - b}}}$$

№ 2.4 Есептеңіз: ${\left( {\dfrac{{{{(a + b)}^{ - n/4}} \cdot {c^{1/2}}}}{{{a^{2 - n}}{b^{ - 3/4}}}}} \right)^{4/3}}:{\left( {\dfrac{{{b^3}{c^4}}}{{{{(a + b)}^{2n}}{a^{16 - 8n}}}}} \right)^{1/6}};b = 0,04$

Шешуі: ММЖ: $a \ne - b = - 0,04$ $$X = {\left( {\dfrac{{{{(a + b)}^{ - n/4}} \cdot {c^{1/2}}}}{{{a^{2 - n}}{b^{ - 3/4}}}}} \right)^{4/3}} = \dfrac{{{{(a + b)}^{ - n/3}} \cdot {c^{2/3}}}}{{{a^{(8 - 4n)/3}}{b^{ - 1}}}} = \dfrac{{b \cdot {c^{2/3}}}}{{{a^{(8 - 4n)/3}} \cdot {{(a + b)}^{n/3}}}}$$ $$Y = {\left( {\dfrac{{{b^3}{c^4}}}{{{{(a + b)}^{2n}}{a^{16 - 8n}}}}} \right)^{1/6}} = \dfrac{{{b^{1/2}} \cdot {c^{2/3}}}}{{{{(a + b)}^{n/3}} \cdot {a^{(8 - 4n/3)}}}}$$ Онда $$X:Y = \dfrac{{b \cdot {c^{2/3}}}}{{{a^{(8 - 4n)/3}} \cdot {{(a + b)}^{n/3}}}}:\dfrac{{{b^{1/2}} \cdot {c^{2/3}}}}{{{{(a + b)}^{n/3}} \cdot {a^{(8 - 4n)/3}}}} = $$ $$ = \dfrac{{b \cdot {c^{2/3}} \cdot {{(a + b)}^{n/3}} \cdot {a^{(8 - 4n)/3}}}}{{{a^{(8 - 4n)/3}} \cdot {{(a + b)}^{n/3}} \cdot {b^{1/2}} \cdot {c^{2/3}}}} = {b^{1/2}} = {(0,04)^{1/2}} = \sqrt {0,04} = 0,2.$$

№ 2.5 Өрнекті ықшамдаңыз: $\dfrac{{2{x^{ - 1/3}}}}{{{x^{2/3}} - 3{x^{ - 1/3}}}} - \dfrac{{{x^{2/3}}}}{{{x^{5/3}} - {x^{2/3}}}} - \dfrac{{x + 1}}{{{x^2} - 4x + 3}}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{x \ne 0}\\{x \ne 1}\\{x \ne 3}\end{array}} \right.$ $${\dfrac{{2{x^{ - 1/3}}}}{{{x^{2/3}} - 3{x^{ - 1/3}}}} - \dfrac{{{x^{2/3}}}}{{{x^{5/3}} - {x^{2/3}}}} - \dfrac{{x + 1}}{{{x^2} - 4x + 3}} = \dfrac{{2{x^{ - 1/3}}}}{{{x^{ - 1/3}}(x - 3)}} - }$$ $${ - \dfrac{{{x^{2/3}}}}{{{x^{2/3}}(x - 1)}} - \dfrac{{x + 1}}{{(x - 1)(x - 3)}} = \dfrac{2}{{x - 3}} - \dfrac{1}{{x - 1}} - \dfrac{{x + 1}}{{(x - 1)(x - 3)}} = }$$ $${ = \dfrac{{2x - 2 - x + 3 - x - 1}}{{(x - 1)(x - 3)}} = \dfrac{0}{{(x - 1)(x - 3)}} = 0}$$

№ 2.6 Өрнекті ықшамдаңыз: $\dfrac{{{{(\sqrt a + \sqrt b )}^2} - 4b}}{{(a - b):\left( {\sqrt {\dfrac{1}{b}} + 3\sqrt {\dfrac{1}{a}} } \right)}}:\dfrac{{a + 9b + 6\sqrt {ab} }}{{\dfrac{1}{{\sqrt b }} + \dfrac{1}{{\sqrt a }}}}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{a \ne b}\\{a \gt 0}\\{b \gt 0}\end{array}} \right.$ $$\dfrac{{{{(\sqrt a + \sqrt b )}^2} - 4b}}{{(a - b):\left( {\sqrt {\dfrac{1}{b}} + 3\sqrt {\dfrac{1}{a}} } \right)}}:\dfrac{{a + 9b + 6\sqrt {ab} }}{{\dfrac{1}{{\sqrt b }} + \dfrac{1}{{\sqrt a }}}} = \dfrac{{a + 2\sqrt {ab} + b - 4b}}{{(\sqrt a - \sqrt b )(\sqrt a + \sqrt b ):\left( {\dfrac{1}{{\sqrt b }} + \dfrac{3}{{\sqrt a }}} \right)}} \div $$ $$\dfrac{{{{(\sqrt a + 3\sqrt b )}^2}}}{{\dfrac{{\sqrt a + \sqrt b }}{{\sqrt {ab} }}}} = \dfrac{{a + 2\sqrt {ab} - 3b}}{{(\sqrt a - \sqrt b )(\sqrt a + \sqrt b ):\dfrac{{\sqrt a + 3\sqrt b }}{{\sqrt {ab} }}}}:\dfrac{{{{(\sqrt a + 3\sqrt b )}^2}\sqrt {ab} }}{{(\sqrt a + \sqrt b )}} = $$ $${ = \dfrac{{(a + 2\sqrt {ab} - 3b)(\sqrt a + 3\sqrt b )(\sqrt a + \sqrt b )}}{{(\sqrt a - \sqrt b )(\sqrt a + \sqrt b )\sqrt {ab} {{(\sqrt a + 3\sqrt b )}^2}\sqrt {ab} }} = }$$ $${ = \dfrac{{a + 2\sqrt {ab} - 3b}}{{ab(a - \sqrt {ab} + 3\sqrt {ab - 3b} )}} = \dfrac{1}{{ab}}}$$

№ 2.7 Өрнекті ықшамдаңыз: $\dfrac{{{{(\sqrt[4]{m} + \sqrt[4]{n})}^2} + {{(\sqrt[4]{m} - \sqrt[4]{n})}^2}}}{{2(m - n)}}:\dfrac{1}{{\sqrt {{m^3}} - \sqrt {{n^3}} }} - 3\sqrt {mn} $

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{m \ne n}\\{m \gt 0}\\{n \gt 0.}\end{array}} \right.$ $${\dfrac{{{{(\sqrt[4]{m} + \sqrt[4]{n})}^2} + {{(\sqrt[4]{m} - \sqrt[4]{n})}^2}}}{{2(m - n)}}:\dfrac{1}{{\sqrt {{m^3}} - \sqrt {{n^3}} }} - 3\sqrt {mn} = }$$ $${ = \dfrac{{\sqrt m + 2\sqrt[4]{{mn}} + \sqrt n + \sqrt m - 2\sqrt[4]{{mn}} + \sqrt n }}{{2(\sqrt m - \sqrt n )(\sqrt m + \sqrt n )}} \times }$$ $${ \times \dfrac{{(\sqrt m - \sqrt n )\left( {{{(\sqrt m )}^2} + \sqrt {mn} + {{(\sqrt n )}^2}} \right)}}{1} - 3\sqrt {mn} = }$$ $${ = \dfrac{{2(\sqrt m + \sqrt n )(\sqrt m - \sqrt n )\left( {{{(\sqrt m )}^2} + \sqrt {mn} + {{(\sqrt n )}^2}} \right)}}{{2(\sqrt m - \sqrt n )(\sqrt m + \sqrt n )}} - 3\sqrt {mn} = }$$ $${ = {{(\sqrt m )}^2} + \sqrt {mn} + {{(\sqrt n )}^2} - 3\sqrt {mn} = {{(\sqrt m )}^2} - }$$ $${ - 2\sqrt {mn} + {{(\sqrt n )}^2} = {{(\sqrt m - \sqrt n )}^2}}$$

№ 2.8 Өрнекті ықшамдаңыз: ${\left( {\left( {\dfrac{{{2^{3/2}} + 27{y^{3/5}}}}{{\sqrt 2 + 3\sqrt[5]{y}}} + 3\sqrt[{10}]{{32{y^2}}} - 2} \right) \cdot {3^{ - 2}}} \right)^5}$

Шешуі: ММЖ: $\sqrt 2 + 3\sqrt[5]{y} \ne 0, \Leftrightarrow y \ne - {\left( {\dfrac{{\sqrt 2 }}{3}} \right)^5}$ $${\left( {\left( {\dfrac{{{2^{3/2}} + 27{y^{3/5}}}}{{\sqrt 2 + 3\sqrt[5]{y}}} + 3\sqrt[{10}]{{32{y^2}}} - 2} \right) \cdot {3^{ - 2}}} \right)^5} = $$ $$ = {\left( {\left( {\dfrac{{{{(\sqrt 2 )}^3} + {{(3\sqrt[5]{y})}^3}}}{{\sqrt 2 + 3\sqrt[5]{y}}} + 3\sqrt 2 \sqrt[5]{y} - 2} \right) \cdot \dfrac{1}{9}} \right)^5} = $$ $$ = {\left( {\left( {\dfrac{{\left. {(\sqrt 2 + 3\sqrt[5]{y}){{(\sqrt 2 )}^2} - 3\sqrt 2 \sqrt[5]{y} + {{(3\sqrt[5]{y})}^2}} \right)}}{{\sqrt 2 + 3\sqrt[5]{y}}} + 3\sqrt 2 \sqrt[5]{y} - 2} \right) \cdot \dfrac{1}{9}} \right)^5} = $$ $$ = {\left( {\left( {2 + 9\sqrt[5]{{{y^2}}} - 2} \right) \cdot \dfrac{1}{9}} \right)^5} = {\left( {9\sqrt[5]{{{y^2}}} \cdot \dfrac{1}{9}} \right)^5} = {\left( {\sqrt[5]{{{y^2}}}} \right)^5} = {y^2}$$

№ 2.9 Өрнекті ықшамдаңыз: ${\dfrac{{2\sqrt {1 + \dfrac{1}{4}{{\left( {\sqrt {\dfrac{1}{t}} - \sqrt t } \right)}^2}} }}{{\sqrt {1 + \dfrac{1}{4}{{\left( {\sqrt {\dfrac{1}{t}} - \sqrt t } \right)}^2}} - \dfrac{1}{2}\left( {\sqrt {\dfrac{1}{t}} - \sqrt t } \right)}}}$

Шешуі: ММЖ: $0 \lt t \ne 1$ $${\dfrac{{2\sqrt {1 + \dfrac{1}{4}{{\left( {\sqrt {\dfrac{1}{t}} - \sqrt t } \right)}^2}} }}{{\sqrt {1 + \dfrac{1}{4}{{\left( {\sqrt {\dfrac{1}{t}} - \sqrt t } \right)}^2}} - \dfrac{1}{2}\left( {\sqrt {\dfrac{1}{t}} - \sqrt t } \right)}} = \dfrac{{2\sqrt {1 + \dfrac{1}{4}\left( {\dfrac{1}{t} - 2 + t} \right)} }}{{\sqrt {1 + \dfrac{1}{4}\left( {\dfrac{1}{t} - 2 + t} \right) - \dfrac{1}{2}\left( {\dfrac{1}{{\sqrt t }} - \sqrt t } \right)} }} = }$$ $${ = \dfrac{{2\sqrt {1 + \dfrac{1}{4} \cdot \dfrac{{1 - 2t + {t^2}}}{t}} }}{{\sqrt {1 + \dfrac{1}{4} \cdot \dfrac{{1 - 2t + {t^2}}}{t}} - \dfrac{1}{2} \cdot \dfrac{{1 - t}}{{\sqrt t }}}} = \dfrac{{2\sqrt {\dfrac{{4t + 1 - 2t + {t^2}}}{{4t}}} }}{{\sqrt {\dfrac{{4t + 1 - 2t + {t^2}}}{{4t}}} - \dfrac{{1 - t}}{{2\sqrt t }}}} = }$$ $${ = \dfrac{{\sqrt {\dfrac{{1 + 2t + {t^2}}}{t}} }}{{\dfrac{1}{2}\sqrt {\dfrac{{1 + 2t + {t^2}}}{t}} - \dfrac{{1 - t}}{{2\sqrt t }}}} = \dfrac{{\dfrac{{1 + t}}{{\sqrt t }}}}{{\dfrac{{1 + t}}{{2\sqrt t }} - \dfrac{{1 - t}}{{2\sqrt t }}}} = \dfrac{{\dfrac{{1 + t}}{{\sqrt t }}}}{{\dfrac{{1 + t - 1 + t}}{{2\sqrt t }}}} = \dfrac{{\dfrac{{1 + t}}{{\sqrt t }}}}{{\dfrac{{2t}}{{2\sqrt t }}}} = \dfrac{{1 + t}}{t}}$$

№ 2.10 Өрнекті ықшамдаңыз: $\dfrac{{1 + \dfrac{2}{{\sqrt {t + 4} }}}}{{2 - \sqrt {t + 4} }} + \sqrt {t + 4} + \dfrac{4}{{\sqrt {t + 4} }}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{t + 4 \gt 0,}\\{2 - \sqrt {t + 4} \ne 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{lllllllllllllll}{t \gt - 4,}\\{t \ne 0.}\end{array}} \right.} \right.$ $$1 + \dfrac{2}{{\sqrt {t + 4} }} + \sqrt {t + 4} + \dfrac{4}{{\sqrt {t + 4} }} = t \cdot \dfrac{{\dfrac{{\sqrt {t + 4} + 2}}{{\sqrt {t + 4} }}}}{{2 - \sqrt {t + 4} }} + \dfrac{{{{(\sqrt {t + 4} )}^2} + 4}}{{\sqrt {t + 4} }} = $$ $$ = t \cdot \dfrac{{\sqrt {t + 4} + 2}}{{\sqrt {t + 4} (2 - \sqrt {t + 4} }} + \dfrac{{{{(\sqrt {t + 4} )}^2} + 4}}{{\sqrt {t + 4} }} = n$$ $${ = \dfrac{{t(\sqrt {t + 4} + 2)(2 + \sqrt {t + 4} )}}{{\sqrt {t + 4} (2 - \sqrt {t + 4} )(2 + \sqrt {t + 4} )}} + \dfrac{{{{(\sqrt {t + 4} )}^2} + 4}}{{\sqrt {t + 4} }} = }$$ $${ = \dfrac{{t{{(\sqrt {t + 4} + 2)}^2}}}{{\sqrt {t + 4} (4 - t - 4)}} + \dfrac{{{{(\sqrt {t + 4} )}^2} + 4}}{{\sqrt {t + 4} }} = }$$ $${ = \dfrac{{ - (t + 4 + 4\sqrt {t + 4} + 4)}}{{\sqrt {t + 4} }} + \dfrac{{t + 4 + 4}}{{\sqrt {t + 4} }} = }$$ $${ = \dfrac{{ - t - 4\sqrt {t + 4} - 8 + t + 8}}{{\sqrt {t + 4} }} = - \dfrac{{4\sqrt {t + 4} }}{{\sqrt {t + 4} }} = - 4}$$

№ 2.11 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{1 + \sqrt x }}{{\sqrt {1 + x} }} - \dfrac{{\sqrt {1 + x} }}{{1 + \sqrt x }}} \right)^2} - {\left( {\dfrac{{1 - \sqrt x }}{{\sqrt {1 + x} }} - \dfrac{{\sqrt {1 + x} }}{{1 - \sqrt x }}} \right)^2}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{1 + x \ge 0,}\\{x \ge 0,}\\{1 - \sqrt x \ne 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{lllllllllllllll}{x \ge 0}\\{x \ne 1}\end{array}} \right.} \right.$ $${{{\left( {\dfrac{{1 + \sqrt x }}{{\sqrt {1 + x} }} - \dfrac{{\sqrt {1 + x} }}{{1 + \sqrt x }}} \right)}^2} - {{\left( {\dfrac{{1 - \sqrt x }}{{\sqrt {1 + x} }} - \dfrac{{\sqrt {1 + x} }}{{1 - \sqrt x }}} \right)}^2} = {{\left( {\dfrac{{{{(1 + \sqrt x )}^2} - {{(\sqrt {1 + x} )}^2}}}{{\sqrt {1 + x} (1 + \sqrt x )}}} \right)}^2} - }$$ $${ - {{\left( {\dfrac{{{{(1 - \sqrt x )}^2} - {{(\sqrt {1 + x} )}^2}}}{{\sqrt {1 + x(1 - \sqrt x )} }}} \right)}^2} = {{\left( {\dfrac{{1 + 2\sqrt x + x - 1 - x}}{{\sqrt {1 + x} (1 + \sqrt x )}}} \right)}^2} - }$$ $${{{\left. { - {{\left( {\dfrac{{1 - 2\sqrt x + x - 1 - x}}{{\sqrt {1 + x} {{(1 - \sqrt x )}^2}}}} \right)}^2} = \left( {\dfrac{{2\sqrt x }}{{\sqrt {1 + x} (1 + \sqrt x }}} \right)} \right)}^2} - }$$ $${ - {{\left( {\dfrac{{ - 2\sqrt x }}{{\sqrt {1 + x} (1 - \sqrt x )}}} \right)}^2} = \dfrac{{4x}}{{{{(\sqrt {1 + x} (1 + \sqrt x ))}^2}}} - }$$ $${ - \dfrac{{4x}}{{{{(\sqrt {1 + x} (1 - \sqrt x ))}^2}}} = \dfrac{{4x\left( {{{(1 - \sqrt x )}^2} - {{(1 + \sqrt x )}^2}} \right)}}{{{{(\sqrt {1 + x} (1 + \sqrt x )(1 - \sqrt x ))}^2}}} = }$$ $${ = \dfrac{{4x(1 - 2\sqrt x + x - 1 - 2\sqrt x - x)}}{{(1 + x){{(1 - x)}^2}}} = \dfrac{{ - 16x\sqrt x }}{{(1 + x)(1 - x)(1 - x)}} = \dfrac{{16x\sqrt x }}{{\left( {1 - {x^2}} \right)(x - 1)}}}$$

№ 2.12 Өрнекті ықшамдаңыз: $\dfrac{{x - 1}}{{x + {x^{1/2}} + 1}}:\dfrac{{{x^{0,5}} + 1}}{{{x^{1,5}} - 1}} + \dfrac{2}{{{x^{ - 0,5}}}}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{x \gt 0}\\{x \ne 1}\end{array}} \right.$ $${\dfrac{{x - 1}}{{x + {x^{1/2}} + 1}}:\dfrac{{{x^{0,5}} + 1}}{{{x^{1,5}} - 1}} + \dfrac{2}{{{x^{ - 0,5}}}} = \dfrac{{\left( {{x^{1/2}} - 1} \right)\left( {{x^{1/2}} + 1} \right)}}{{x + {x^{1/2}} + 1}} \cdot \dfrac{{{{\left( {{x^{1/2}}} \right)}^3} - 1}}{{{x^{1/2}} + 1}} + \dfrac{2}{{\dfrac{1}{{{x^{1/2}}}}}} = }$$ $${ = \dfrac{{\left( {{x^{1/2}} - 1} \right)\left( {{x^{1/2}} - 1} \right)\left( {x + {x^{1/2}} + 1} \right)}}{{x + {x^{1/2}} + 1}} + 2{x^{1/2}} = {{\left( {{x^{1/2}} - 1} \right)}^2} + 2{x^{1/2}} = x - 2{x^{1/2}} + }$$ $${ + 1 + 2{x^{1/2}} = x + 1.}$$

№ 2.13 Өрнекті ықшамдаңыз: $\left( {\dfrac{1}{{\sqrt a + \sqrt {a + 1} }} + \dfrac{1}{{\sqrt a - \sqrt {a - 1} }}} \right):\left( {1 + \sqrt {\dfrac{{a + 1}}{{a - 1}}} } \right)$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{a \ge 0}\\{\sqrt a - \sqrt {a - 1} \ne 0,}\\{\dfrac{{a + 1}}{{a - 1}} \ge 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{lllllllllllllll}{a \ge 0,}\\{a \gt 1}\end{array} \Leftrightarrow a \gt 1.} \right.} \right.$ X - бірінші жақшадағы ал, Y өрнек болсын. $${X = \dfrac{1}{{\sqrt a + \sqrt {a + 1} }} + \dfrac{1}{{\sqrt a - \sqrt {a - 1} }} = \dfrac{{\sqrt a - \sqrt {a - 1} + \sqrt a + \sqrt {a + 1} }}{{(\sqrt a + \sqrt {a + 1} )(\sqrt a - \sqrt {a - 1} )}} = }$$ $${ = \dfrac{{2\sqrt a + \sqrt {a + 1} - \sqrt {a - 1} }}{{a + \sqrt {a(a + 1)} - \sqrt {a(a - 1)} - \sqrt {(a - 1)(a + 1)} }}{\rm{; }}}$$ $${Y = 1 + \dfrac{{\sqrt {a + 1} }}{{\sqrt {a - 1} }} = \dfrac{{\sqrt {a + 1} + \sqrt {a - 1} }}{{\sqrt {a - 1} }}.}$$ Онда $${X:Y = \dfrac{{2\sqrt a + \sqrt {a + 1} - \sqrt {a - 1} }}{{a + \sqrt {a(a + 1)} - \sqrt {a(a - 1)} - \sqrt {(a - 1)(a + 1)} }}:\dfrac{{\sqrt {a + 1} + \sqrt {a - 1} }}{{\sqrt {a - 1} }} = }$$ $${ = \dfrac{{\sqrt {a - 1} (2\sqrt a + \sqrt {a + 1} - \sqrt {a - 1} )}}{{a\sqrt {a + 1} + a\sqrt a + \sqrt a - a\sqrt {a - 1} - \sqrt {a - 1} + a\sqrt {a - 1} - a\sqrt a + \sqrt a - a\sqrt {a + 1} + \sqrt {a + 1} }} = }$$ $${ = \dfrac{{\sqrt {a - 1} (2\sqrt a + \sqrt {a + 1} - \sqrt {a - 1} )}}{{2\sqrt a + \sqrt {a + 1} - \sqrt {a - 1} }} = \sqrt {a - 1} {\rm{. }}}$$

№ 2.14 Өрнекті ықшамдаңыз: $\dfrac{{x - y}}{{{x^{3/4}} + {x^{1/2}}{y^{1/4}}}} \cdot \dfrac{{{x^{1/2}}{y^{1/4}} + {x^{1/4}}{y^{1/2}}}}{{{x^{1/2}} + {y^{1/2}}}} \cdot \dfrac{{{x^{1/4}}{y^{ - 1/4}}}}{{{x^{1/2}} - 2{x^{1/4}}{y^{1/4}} + {y^{1/2}}}}$

Шешуі: $\left\{ {\begin{array}{lllllllllllllll}{x \ge 0}\\{y \ge 0}\\{x \ne y}\end{array}} \right.$ $${\dfrac{{x - y}}{{{x^{3/4}} + {x^{1/2}}{y^{1/4}}}} \cdot \dfrac{{{x^{1/2}}{y^{1/4}} + {x^{1/4}}{y^{1/2}}}}{{{x^{1/2}} + {y^{1/2}}}} \cdot \dfrac{{{x^{1/4}}{y^{ - 1/4}}}}{{{x^{1/2}} - 2{x^{1/4}}{y^{1/4}} + {y^{1/2}}}} = }$$ $${ = \dfrac{{x - y}}{{\sqrt[4]{{{x^3}}} + \sqrt[4]{{{x^2}}}\sqrt[4]{y}}} \cdot \dfrac{{\sqrt[4]{{{x^2}}}\sqrt[4]{y} + \sqrt[4]{x}\sqrt[4]{{{y^2}}}}}{{\sqrt x + \sqrt y }} \cdot \dfrac{{\sqrt[4]{x}}}{{\sqrt[4]{y}\left( {\sqrt[4]{{{x^2}}} - 2\sqrt[4]{x}\sqrt[4]{y} + \sqrt[4]{{{y^2}}}} \right)}} = }$$ $${ = \dfrac{{x - y}}{{\sqrt[4]{{{x^2}}}(\sqrt[4]{x} + \sqrt[4]{y})}} \cdot \dfrac{{\sqrt[4]{x}\sqrt[4]{y}(\sqrt[4]{x} + \sqrt[4]{y})}}{{\sqrt x + \sqrt y }} \cdot \dfrac{{\sqrt[4]{x}}}{{\sqrt[4]{y}{{(\sqrt[4]{x} - \sqrt[4]{y})}^2}}} = }$$ $${ = \dfrac{{x - y}}{{(\sqrt x + \sqrt y ){{(\sqrt[4]{x} - \sqrt[4]{y})}^2}}} = \dfrac{{(\sqrt x - \sqrt y )(\sqrt x + \sqrt y )}}{{(\sqrt x + \sqrt y ){{(\sqrt[4]{x} - \sqrt[4]{y})}^2}}} = }$$

№ 2.15 Өрнекті ықшамдаңыз: $\sqrt[n]{{{y^{\dfrac{{2n}}{{m - n}}}}}}:\sqrt[m]{{\dfrac{{{{(m - n)}^2} + 4mn}}{{{m^2} - {n^2}}}}}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{m \ne \pm n,}\\{y \ne 0,}\\{y \gt 0{\rm{ }}{\rm{, }}m = 2k.}\end{array}} \right.$ $$\sqrt[n]{{\dfrac{{2n}}{{{y^{m - n}}}}}}:\sqrt[m]{{\dfrac{{{{(m - n)}^2} + 4mn}}{{{m^2} - {n^2}}}}} = {y^{\dfrac{{2n}}{{n(m - n)}}}}:{y^{\dfrac{{{m^2} - 2mn + {n^2} + 4mn}}{{m(m + n)(m - n)}}}} = $$ $$ = {y^{\dfrac{2}{{m - n}}}}:{y^{\dfrac{{{m^2} + 2mn + {n^2}}}{{m(m + n)(m - n)}}}} = {y^{\dfrac{2}{{m - n}}}}:{y^{\dfrac{{{{(m + n)}^2}}}{{m(m + n)(m - n)}}}} = {y^{\dfrac{2}{{m - n}}}}:{y^{\dfrac{{m + n}}{{m(m - n)}}}} = $$ $$ = {y^{\dfrac{2}{{m - n}} - \dfrac{{m + n}}{{m(m - n)}}}} = {y^{\dfrac{{2m - m - n}}{{m(m - n)}}}} = {y^{\dfrac{{m - n}}{{m(m - n)}}}} = {y^{\dfrac{1}{m}}} = \sqrt[m]{y}.$$

 
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