Иррационал өрнектер (Сканави (А) 2.1-2.15)

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

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

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 $${X:Y = \dfrac{{a – b}}{{\sqrt {ab} }}:\dfrac{{{{(\sqrt a – \sqrt b )}^2}}}{{\sqrt {ab} }} = \dfrac{{(a – b) \cdot \sqrt {ab} }}{{\sqrt {ab} \cdot {{(\sqrt a – \sqrt b )}^2}}} = \dfrac{{a – b}}{{{{(\sqrt a – \sqrt b )}^2}}} = }$$ $${ = \dfrac{{(a – b){{(\sqrt a + \sqrt b )}^2}}}{{{{(\sqrt a – \sqrt b )}^2}{{(\sqrt a + \sqrt b )}^2}}} = \dfrac{{(a – b){{(\sqrt a + \sqrt b )}^2}}}{{{{((\sqrt a – \sqrt b )(\sqrt a + \sqrt b ))}^2}}} = \dfrac{{(a – b){{(\sqrt a + \sqrt b )}^2}}}{{{{(a – b)}^2}}} = }$$ $${ = \dfrac{{{{(\sqrt a + \sqrt b )}^2}}}{{a – b}}}$$

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 $$X:Y = \dfrac{{b \cdot {c^{2/3}}}}{{{a^{(8 – 4n)/3}} \cdot {{(a + b)}^{n/3}}}}:\dfrac{{{b^{1/2}} \cdot {c^{2/3}}}}{{{{(a + b)}^{n/3}} \cdot {a^{(8 – 4n)/3}}}} = $$ $$ = \dfrac{{b \cdot {c^{2/3}} \cdot {{(a + b)}^{n/3}} \cdot {a^{(8 – 4n)/3}}}}{{{a^{(8 – 4n)/3}} \cdot {{(a + b)}^{n/3}} \cdot {b^{1/2}} \cdot {c^{2/3}}}} = {b^{1/2}} = {(0,04)^{1/2}} = \sqrt {0,04} = 0,2.$$

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

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

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

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

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

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

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

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

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 $${X:Y = \dfrac{{2\sqrt a + \sqrt {a + 1} – \sqrt {a – 1} }}{{a + \sqrt {a(a + 1)} – \sqrt {a(a – 1)} – \sqrt {(a – 1)(a + 1)} }}:\dfrac{{\sqrt {a + 1} + \sqrt {a – 1} }}{{\sqrt {a – 1} }} = }$$ $${ = \dfrac{{\sqrt {a – 1} (2\sqrt a + \sqrt {a + 1} – \sqrt {a – 1} )}}{{a\sqrt {a + 1} + a\sqrt a + \sqrt a – a\sqrt {a – 1} – \sqrt {a – 1} + a\sqrt {a – 1} – a\sqrt a + \sqrt a – a\sqrt {a + 1} + \sqrt {a + 1} }} = }$$ $${ = \dfrac{{\sqrt {a – 1} (2\sqrt a + \sqrt {a + 1} – \sqrt {a – 1} )}}{{2\sqrt a + \sqrt {a + 1} – \sqrt {a – 1} }} = \sqrt {a – 1} {\rm{. }}}$$

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

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