Өрнектерді ықшамдаңыз.
№ 1 Өрнекті ықшамдаңыз: ${\dfrac{{{m^4} – 49}}{{{m^2} + 7}} – \dfrac{{{m^6} – 343}}{{{m^4} + 7{m^2} + 49}}}$
Шешуі: $$ = \dfrac{{\left( {{m^2} – 7} \right)\left( {{m^2} + 7} \right)}}{{{m^2} + 7}} – \dfrac{{\left( {{m^2} – 7} \right)\left( {{m^4} + 7{m^2} + 49} \right)}}{{{m^4} + 7{m^2} + 49}}$$ $${ = {m^2} – 7 – \left( {{m^2} – 7} \right) = {m^2} – 7 – {m^2} + 7 = 0}$$
№ 2 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{1 + n}}{{{n^2} – mn}} – \dfrac{{1 – m}}{{{m^2} – mn}}} \right) \cdot {{\left( {\dfrac{{m + n}}{{{m^2}n – {n^2}m}}} \right)}^{ – 1}}}$
Шешуі: $${ = \left( {\dfrac{{1 + n}}{{n(n – m)}} – \dfrac{{1 – m}}{{ – m(n – m)}}} \right) \cdot \left( {\dfrac{{{m^2}n – {n^2}m}}{{m + n}}} \right) = }$$ $$ = \left( {\frac{{1 + n}}{{n\left( {n – m} \right)}} + \frac{{1 – m}}{{m\left( {n – m} \right)}}} \right) \cdot \left( {\frac{{{m^2}n – {n^2}m}}{{m + n}}} \right) = $$ $$ = \dfrac{{m + mn + n – mn}}{{mn(n – m)}} \cdot \dfrac{{ – mn(n – m)}}{{m + n}} = $$ $$ = \dfrac{{m + n}}{{mn \cdot (n – m)}} \cdot \left( { – \dfrac{{mn\left( {n – m} \right)}}{{m + n}}} \right) = – 1$$
№ 3 Өрнекті ықшамдаңыз: $\left( {\dfrac{{ab}}{{a – b}} + a} \right) \cdot \left( {\dfrac{{ab}}{{a + b}} – a} \right):\dfrac{{{a^2}{b^2}}}{{{b^2} – {a^2}}}$
Шешуі: $$ = \dfrac{{ab + {a^2} – ab}}{{a – b}} \cdot \dfrac{{ab – {a^2} – ab}}{{a + b}} \cdot \dfrac{{{b^2} – {a^2}}}{{a^2 {b^2}}} = $$ $${\dfrac{{{a^2}}}{{a – b}} \cdot \dfrac{{ – {a^2}}}{{a + b}} \cdot \dfrac{{{b^2} – {a^2}}}{{a^2 {b^2}}} = – \dfrac{{{a^4}}}{{{a^2} – {b^2}}} \cdot \dfrac{{{b^2} – {a^2}}}{{{a^2}{b^2}}} = }$$ $${ = \dfrac{{{a^4}}}{{{b^2} – {a^2}}} \cdot \dfrac{{{b^2} – {a^2}}}{{{a^2}{b^2}}} = \dfrac{{{a^2}}}{{{b^2}}} = {{\left( {\dfrac{a}{b}} \right)}^2}}$$
№ 4 Егер $b=-0,5$ болса, $\dfrac{{{{(a + b)}^2} – {{(ab + 1)}^2}}}{{{a^2} – 1}}:0,75$ есептеңіз.
Шешуі: $$ = \dfrac{{{a^2} + 2ab + {b^2} – {a^2}{b^2} – 2ab – 1}}{{{a^2} – 1}}:\dfrac{3}{4} = $$ $$ = \dfrac{{{a^2} + {b^2} – {a^2}{b^2} – 1}}{{{a^2} – 1}}:\dfrac{3}{4} = \dfrac{{{a^2} – {a^2}{b^2} + {b^2} – 1}}{{{a^2} – 1}} \cdot \dfrac{4}{3} = $$ $$ = \dfrac{{{a^2}\left( {1 – {b^2}} \right) – \left( {1 – {b^2}} \right)}}{{{a^2} – 1}} \cdot \dfrac{4}{3} = \dfrac{{\left( {1 – {b^2}} \right)\left( {{a^2} – 1} \right)}}{{{a^2} – 1}} \cdot \dfrac{4}{3} = $$ $$ = \left( {1 – {b^2}} \right) \cdot \dfrac{4}{3} = \left( {1 – {{\left( { – \dfrac{1}{2}} \right)}^2}} \right) \cdot \dfrac{4}{3} = \left( {1 – \dfrac{1}{4}} \right) \cdot \dfrac{4}{3} = \dfrac{3}{4} \cdot \dfrac{4}{3} = 1$$
№ 5 Өрнекті ықшамдаңыз: $\left( {\dfrac{{2ab}}{{{a^2} – {b^2}}} + \dfrac{{a – b}}{{2(a + b)}}} \right) \cdot \dfrac{{2a}}{{a + b}} – \dfrac{b}{{a – b}}$
Шешуі: $$ = \left( {\dfrac{{2ab}}{{(a – b)(a + b)}} + \dfrac{{a – b}}{{2(a + b)}}} \right) \cdot \dfrac{{2a}}{{a + b}} – \dfrac{b}{{a – b}} = $$ $$ = \dfrac{{4ab + {a^2} – 2ab + {b^2}}}{{2(a – b)(a + b)}} \cdot \dfrac{{2a}}{{a + b}} – \dfrac{b}{{a – b}} = $$ $$ = \dfrac{{{a^2} + 2ab + {b^2}}}{{2(a – b)(a + b)}} \cdot \dfrac{{2a}}{{a + b}} – \dfrac{b}{{a – b}} = $$ $$ = \dfrac{{{{(a + b)}^2}}}{{2(a – b)(a + b)}} \cdot \dfrac{{2a}}{{a + b}} – \dfrac{b}{{a – b}} = $$ $$ = \dfrac{a}{{a – b}} – \dfrac{b}{{a – b}} = \dfrac{{a – b}}{{a – b}} = 1$$
№ 6 Егер $x=3$ болса,${\dfrac{{3x + 2}}{{2x + 3}} – \dfrac{{4x – 1}}{{2x + 3}} + \dfrac{{2{x^2} + 3x}}{{4{x^2} + 12x + 9}}}$ есептеңіз.
Шешуі: $$ = \dfrac{{3x + 2 – 4x + 1}}{{2x + 3}} + \dfrac{{x(2x + 3)}}{{{{(2x + 3)}^2}}} = $$ $$ = \dfrac{{3 – x}}{{2x + 3}} + \dfrac{x}{{2x + 3}} = \dfrac{3}{{2x + 3}}$$ $$x = 3\quad \Rightarrow \quad \dfrac{3}{{2 \cdot 3 + 3}} = \dfrac{3}{9} = \dfrac{1}{3}$$
№ 7 Өрнекті ықшамдаңыз: $\dfrac{6}{{7(x – 3)}} – \dfrac{1}{{{x^2} – 6x + 9}} + \dfrac{1}{{{x^2} – 9}}$
Шешуі: $$ = \dfrac{6}{{7(x – 3)}} – \dfrac{1}{{{{(x – 3)}^2}}} + \dfrac{1}{{(x – 3)(x + 3)}} = $$ $$ = \dfrac{{6{x^2} – 54 – 7x – 21 + 7x – 21}}{{7{{(x – 3)}^2}(x + 3)}} = \dfrac{{6{x^2} – 96}}{{7{{(x – 3)}^2}(x + 3)}} = $$ $$ = \dfrac{{6\left( {{x^2} – 16} \right)}}{{7{{(x – 3)}^2}(x + 3)}}$$ $$x = 4\quad \Rightarrow \quad = \dfrac{{6\left( {{4^2} – 16} \right)}}{{7{{(4 – 3)}^2}\left( {4 + 3} \right)}} = 0$$
№ 8 Өрнекті ықшамдаңыз: $\dfrac{{3a – 4}}{{a + 1}} + \dfrac{a}{{a + 1}}:\dfrac{a}{{{a^2} – 1}} + \dfrac{{5 – 2a}}{{a + 1}} = $
Шешуі: $$ = \dfrac{{3a – 4}}{{a + 1}} + \dfrac{a}{{a + 1}} \cdot \dfrac{{(a + 1)(a – 1)}}{a} + \dfrac{{5 – 2a}}{{a + 1}} = $$ $$ = \dfrac{{3a – 4 + 5 – 2a}}{{a + 1}} + \dfrac{{a – 1}}{1} = \dfrac{{a + 1}}{{a + 1}} + a – 1 = $$ $$ = 1 + a – 1 = a$$
№ 9 Өрнекті ықшамдаңыз: $\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} – \dfrac{1}{{ab}}} \right):\left( {\dfrac{{{a^2}}}{b} + \dfrac{{{b^2}}}{a}} \right)$
Шешуі: $$ = \dfrac{{{a^2} – ab + {b^2}}}{{{a^2}{b^2}}}:\dfrac{{{a^3} + {b^3}}}{{ab}} = $$ $${ = \dfrac{{{a^2} – ab + {b^2}}}{{{a^2}{b^2}}} \cdot \dfrac{{ab}}{{(a + b)\left( {{a^2} – ab + {b^2}} \right)}} = \dfrac{1}{{ab(a + b)}}}$$
№ 10 Өрнекті ықшамдаңыз: $\left( {\dfrac{{{m^3} + {n^3}}}{{{m^2} – {n^2}}} – \dfrac{{{m^2} – {n^2}}}{{m + n}}} \right) \cdot {(mn)^{ – 1}}$
Шешуі: $$ = \left( {\dfrac{{(m + n)\left( {{m^2} – mn + {n^2}} \right)}}{{(m + n)(m – n)}} – \dfrac{{(m + n)(m – n)}}{{m + n}}} \right) \cdot \dfrac{1}{{mn}} = $$ $${ = \left( {\dfrac{{{m^2} – mn + {n^2}}}{{m – n}} – m + n} \right) \cdot \dfrac{1}{{mn}} = }$$ $$ = \dfrac{{{m^2} – mn + {n^2} – {{(m – n)}^2}}}{{m – n}} \cdot \dfrac{1}{{mn}} = $$ $$ = \dfrac{{{m^2} – mn + {n^2} – {m^2} + 2mn – {n^2}}}{{m – n}} \cdot \dfrac{1}{{mn}} = $$ $${ = \dfrac{{mn}}{{m – n}} \cdot \dfrac{1}{{mn}} = \dfrac{1}{{m – n}}}$$
№ 11 Өрнекті ықшамдаңыз: ${\left( {\dfrac{6}{{a – 3}} – \dfrac{3}{{a – 1}} – \dfrac{{a + 1}}{{{a^2} – 4a + 3}}} \right) \cdot {{\left( {\dfrac{{a + 1}}{{a – 3}}} \right)}^{ – 1}}}$
Шешуі: $${ = \left( {\dfrac{6}{{a – 1}} – \dfrac{3}{{a – 1}} – \dfrac{{a + 1}}{{(a – 3)(a – 1)}}} \right) \cdot \dfrac{{a – 3}}{{a + 1}} = }$$ $${ = \dfrac{{6a – 6 – 3a + 9 – a – 1}}{{(a – 3)(a – 1)}} \cdot \dfrac{{a – 3}}{{a + 1}} = \dfrac{{2a + 2}}{{(a – 3)(a – 1)}} \cdot \dfrac{{a – 3}}{{a + 1}} = }$$ $${ = \dfrac{{2(a + 1)}}{{a – 1}} \cdot \dfrac{1}{{(a + 1)}} = \dfrac{2}{{a – 1}}}$$
№ 12 Өрнекті ықшамдаңыз: ${\dfrac{{3{a^2} + 2ax – {x^2}}}{{(3x + a)(a + x)}} + 10 \cdot \dfrac{{ax – 3{x^2}}}{{{a^2} – 9{x^2}}} = }$
Шешуі: $${ = \dfrac{{ – (x – 3a)(x + a)}}{{(3x + a)(a + x)}} + 10 \cdot \dfrac{{x(a – 3x)}}{{(a – 3x)(a + 3x)}} = }$$ $${ = \dfrac{{3a – x}}{{3x + a}} + 10 \cdot \dfrac{x}{{3x + a}} = \dfrac{{3a – x + 10x}}{{3x + a}} = }$$ $${ = \dfrac{{9x + 3a}}{{3x + a}} = \dfrac{{3(3x + a)}}{{3x + a}} = 3}$$
№ 13 Өрнекті ықшамдаңыз: $\dfrac{{{{(a + b)}^2} – 3ab}}{{{a^3} + {a^2}b + a{b^2} + {b^3}}} \cdot \dfrac{{{a^4} – {b^4}}}{{{a^3} + {b^3}}} = $
Шешуі: $$ = \dfrac{{{a^2} + 2ab + {b^2} – 3ab}}{{{a^2}(a + b) + {b^2}(a + b)}} \cdot \dfrac{{\left( {{a^2} – {b^2}} \right)\left( {{a^2} + {b^2}} \right)}}{{(a + b)\left( {{a^2} – ab + {b^2}} \right)}} = $$ $$ = \dfrac{{{a^2} – ab + {b^2}}}{{(a + b)\left( {a^2 + {b^2}} \right)}} \cdot \dfrac{{(a – b)(a + b)\left( {{a^2} + {b^2}} \right)}}{{(a + b)\left( {{a^2} – ab + {b^2}} \right)}} = \dfrac{{a – b}}{{a + b}}$$
№ 14 Өрнекті ықшамдаңыз: $\dfrac{{{m^2} – 9}}{{{m^2} – 1}}:\dfrac{{{m^2} + 4m + 3}}{{{m^2} – 4m + 3}}$
Шешуі: $$ = \dfrac{{(m – 3)(m + 3)}}{{(m – 1)(m + 1)}}:\dfrac{{(m + 1)(m + 3)}}{{(m – 1)(m – 3)}} = $$ $$ = \dfrac{{(m – 3)(m + 3)}}{{(m – 1)(m + 1)}} \cdot \dfrac{{(m – 1)(m – 3)}}{{(m + 1)(m + 3)}} = $$ $$ = \dfrac{{{{(m – 3)}^2}}}{{{{(m + 1)}^2}}} = {\left( {\dfrac{{m – 3}}{{m + 1}}} \right)^2}$$
№ 15 Өрнекті ықшамдаңыз: $\dfrac{x}{{{x^2} + {y^2}}} – \dfrac{{y{{(x – y)}^2}}}{{{x^4} – {y^4}}}$
Шешуі: $$ = \dfrac{x}{{{x^2} + {y^2}}} – \dfrac{{y{{(x – y)}^2}}}{{\left( {{x^2} + {y^2}} \right)\left( {{x^2} – {y^2}} \right)}} = $$ $$ = \dfrac{x}{{{x^2} + {y^2}}} – \dfrac{{y{{(x – y)}^2}}}{{\left( {{x^2} + {y^2}} \right)(x – y)(x + y)}} = $$ $$ = \dfrac{{{x^2} + xy – xy + {y^2}}}{{\left( {{x^2} + {y^2}} \right)(x + y)}} = \dfrac{{{x^2} + {y^2}}}{{\left( {{x^2} + {y^2}} \right)(x + y)}} = \dfrac{1}{{x + y}}$$
№ 16 Егер $a=2,71 \quad b=1,29$ болса, $\dfrac{a^4-b^4}{(a+b)^2-2ab}$ есептеңіз.
Шешуі: $$\dfrac{{{a^4} – {b^4}}}{{{{(a + b)}^2} – 2ab}} = \dfrac{{\left( {{a^2} + {b^2}} \right)\left( {{a^2} – {b^2}} \right)}}{{{a^2} + 2a \cdot b + {b^2} – 2ab}} = $$ $$ = \dfrac{{\left( {{a^2} + {b^2}} \right)\left( {{a^2} – {b^2}} \right)}}{{{a^2} + {b^2}}} = {a^2} – {b^2} = $$ $$ = (a – b)(a + b) = \left| {\left| {\begin{array}{*{20}{l}}{a = 2,71}\\{b = 1,29}\end{array}} \right|} \right| = $$ $$=(2,71 – 1,29)(2,71 + 1,29) = 4 \cdot 1,42 = 5,68$$
№ 17 Өрнекті ықшамдаңыз: $\dfrac{{{x^4} – 4}}{{{x^2} + 2}} + 2\left( {x + \dfrac{3}{2}} \right)$
Шешуі: $$ = \dfrac{{\left( {{x^2} + 2} \right)\left( {{x^2} – 2} \right)}}{{{x^2} + 2}} + 2x + 3 = $$ $${ = {x^2} – 2 + 2x + 3 = {x^2} + 2x + 1 = {{(x + 1)}^2}}$$
№ 18 Өрнекті ықшамдаңыз: $\left( {a + 1 + \dfrac{1}{{a – 1}}} \right):\dfrac{{{a^2}}}{{1 – 2a + {a^2}}}$
Шешуі: $$ = \dfrac{{{a^2} – 1 + 1}}{{a – 1}}:\dfrac{{{a^2}}}{{{{(a – 1)}^2}}} = \dfrac{{{a^2}}}{{a – 1}} \cdot \dfrac{{{{(a – 1)}^2}}}{{{a^2}}} = a – 1$$
№ 19 Өрнекті ықшамдаңыз: $\left( {\dfrac{b}{{{a^2} – ab}} – \dfrac{1}{{a – b}}} \right):\left( {\dfrac{{a + b}}{{{a^2} – ab}} – \dfrac{b}{{ab – {b^2}}}} \right)$
Шешуі: $$ = \left( {\dfrac{b}{{a(a – b)}} – \dfrac{1}{{a – b}}} \right):\left( {\dfrac{{a + b}}{{a(a – b)}} – \dfrac{b}{{b(a – b)}}} \right) = $$ $$ = \dfrac{{b – a}}{{a(a – b)}}:\dfrac{{a + b – a}}{{a(a – b)}} = – \dfrac{1}{a} \cdot \dfrac{{a(a – b)}}{b} = \dfrac{{b – a}}{b}$$
№ 20 Өрнекті ықшамдаңыз: $\dfrac{x}{{x – y}} + \dfrac{{{x^2} + {y^2}}}{{{y^2} – {x^2}}} + \dfrac{x}{{x + y}}$
Шешуі: $$ = \dfrac{x}{{x – y}} – \dfrac{{{x^2} + {y^2}}}{{(x – y)(x + y)}} + \dfrac{{x}}{{x + y}} = $$ $$ = \dfrac{{{x^2} + xy – {x^2} – {y^2} + {x^2} – xy}}{{(x – y)(x + y)}} = \dfrac{{{x^2} – {y^2}}}{{{x^2} – {y^2}}} = 1$$
№ 21 Өрнекті ықшамдаңыз: $\left( {\dfrac{{m – 2}}{{m + 2}} – \dfrac{{m + 2}}{{m – 2}}} \right):\dfrac{{8m}}{{{m^2} – 4}}$
Шешуі: $$ = \dfrac{{{{(m – 2)}^2} – {{(m + 2)}^2}}}{{(m + 2)(m – 2)}} \cdot \dfrac{{(m + 2)(m – 2)}}{{8m}} = $$ $${ = \dfrac{{{m^2} – 4m + 4 – {m^2} – 4m – 4}}{{(m + 2)(m – 2)}} \cdot \dfrac{{\left( {m + 2} \right)\left( {m – 2} \right)}}{{8m}} = \dfrac{{ – 8m}}{{8m}} = – 1}$$
№ 22 Өрнекті ықшамдаңыз: $(a + b) \cdot \left( {\dfrac{1}{a} – \dfrac{1}{b}} \right):\dfrac{{{a^2} – {b^2}}}{{{a^2}{b^2}}}$
Шешуі: $$ = \dfrac{{a + b}}{1} \cdot \dfrac{{b – a}}{{ab}} \cdot \dfrac{{{a^2}{b^2}}}{{{a^2} – {b^2}}} = $$ $${ = \dfrac{{ – \left( {{a^2} – {b^2}} \right)}}{{ab}} \cdot \dfrac{{{a^2}{b^2}}}{{{a^2} – {b^2}}} = – ab}$$
№ 23 Өрнекті ықшамдаңыз: $\left( {\dfrac{{ab}}{{{a^2} – {b^2}}} – \dfrac{b}{{2a – 2b}}} \right):\dfrac{{2b}}{{{a^2} – {b^2}}}$
Шешуі: $$ = \left( {\dfrac{{ab}}{{(a – b)(a + b)}} – \dfrac{b}{{2(a – b)}}} \right):\dfrac{{2b}}{{(a – b)(a + b)}} = $$ $$ = \dfrac{{2ab – ab – {b^2}}}{{2(a – b)(a + b)}} \cdot \dfrac{{(a – b)(a + b)}}{{2b}} = $$ $$ = \dfrac{{ab – {b^2}}}{{4b}} = \dfrac{{b(a – b)}}{{4b}} = \dfrac{{a – b}}{4}$$
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