Өрнектерді ықшамдаңыз.
№ 24 Өрнекті ықшамдаңыз: ${\left( {\dfrac{1}{{2 – 6a}} + \dfrac{1}{{27{a^3} – 1}}:\dfrac{{1 + 3a}}{{1 + 3a + 9{a^2}}}} \right) \cdot \dfrac{{2 + 6a}}{a}}$
Шешуі: $${ = \left( {\dfrac{1}{{ – 2(3a – 1)}} + \dfrac{1}{{(3a – 1)\left( {9{a^2} + 3a + 1} \right)}} \cdot \dfrac{{1 + 3a + 9{a^2}}}{{3a + 1}}} \right) \cdot \dfrac{{2(3a + 1)}}{a} = }$$ $${ = \left( {\dfrac{1}{{ – 2(3a – 1)}} + \dfrac{1}{{(3a – 1)(3a + 1)}}} \right) \cdot \dfrac{{2(3a + 1)}}{a} = }$$ $${ = \dfrac{{3a + 1 – 2}}{{ – 2(3a – 1)(3a + 1)}} \cdot \dfrac{{2(3a + 1)}}{a} = \dfrac{{3a – 1}}{{ – 2(3a – 1)}} \cdot \dfrac{2}{a} = – \dfrac{1}{a}}$$
№ 25 Өрнекті ықшамдаңыз: $\dfrac{{8 – {n^3}}}{{2 + n}}:\left( {2 + \dfrac{{{n^2}}}{{2 + n}}} \right) – \dfrac{{{n^2}}}{{n – 2}} \cdot \dfrac{{4 – {n^2}}}{{{n^2} + 2n}}$
Шешуі: $$ = \dfrac{{ – \left( {{n^3} – 8} \right)}}{{n + 2}}:\dfrac{{4 + 2n + {n^2}}}{{n + 2}} – \dfrac{{{n^2}}}{{n – 2}} \cdot \dfrac{{(2 – n)\left( {2 + n} \right)}}{{n(n + 2)}} = $$ $$ = – \dfrac{{(n – 2)\left( {{n^2} + 2n + 4} \right)}}{{n + 2}} \cdot \dfrac{{n + 2}}{{{n^2} + 2n + 4}} + \dfrac{{{n^2}(n – 2)}}{{n(n – 2)}} = $$ $$ = – n + 2 + n = 2$$
№ 26 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{12}}{{5{a^2} + a – 4}} – \dfrac{{a + 1}}{{3(5a – 4)}}} \right) \cdot \left( {\dfrac{{15a – 12}}{{a + 7}}} \right)}$
Шешуі: $${ = \left( {\dfrac{{12}}{{(a + 1)(5a – 4)}} – \dfrac{{a + 1}}{{3(5a – 4)}}} \right) \cdot \dfrac{{3(5a – 4)}}{{a + 7}} = }$$ $$ = \dfrac{{36 – {a^2} – 2a – 1}}{{3(a + 1)(5a – 4)}} \cdot \dfrac{{3(5a – 4)}}{{a + 7}} = \dfrac{{35 – 2a – {a^2}}}{{(a + 1)(a + 7)}} = $$ $$ = \dfrac{{ – (a + 7)(a – 5)}}{{(a + 1)(a + 7)}} = \dfrac{{5 – a}}{{a + 1}}$$
№ 27 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{x + 5}}{{(x – 9)(x + 9)}} + \dfrac{{x + 7}}{{{{(x – 9)}^2}}}} \right) \cdot {{\left( {\dfrac{{x – 9}}{{x + 3}}} \right)}^2} + \dfrac{{7 + x}}{{9 + x}}}$
Шешуі: $${ = \dfrac{{{x^2} – 4x – 45 + {x^2} + 16x + 63}}{{(x + 9){{(x – 9)}^2}}} \cdot \dfrac{{{{(x – 9)}^2}}}{{{{(x + 3)}^2}}} + \dfrac{{7 + x}}{{9 + x}} = }$$ $${ = \dfrac{{2{x^2} + 12x + 18}}{{x + 9}} \cdot \dfrac{1}{{{{(x + 3)}^2}}} + \dfrac{{7 + x}}{{9 + x}} = }$$ $${ = \dfrac{{2\left( {{x^2} + 6x + 9} \right)}}{{x + 9}} \cdot \dfrac{1}{{{{(x + 3)}^2}}} + \dfrac{{7 + x}}{{9 + x}} = \dfrac{{2{{(x + 3)}^2}}}{{x + 9}} \cdot \dfrac{1}{{{{(x + 3)}^2}}} + \dfrac{{7 + x}}{{9 + x}} = }$$ $${ = \dfrac{2}{{x + 9}} + \dfrac{{7 + x}}{{9 + x}} = \dfrac{{2 + 7 + x}}{{x + 9}} = \dfrac{{x + 9}}{{x + 9}} = 1}$$
№ 28 $a=-2,5; \quad b=0,5$ болғанда $\left( {\dfrac{{{a^2}}}{{a + b}} – \dfrac{{{a^3}}}{{{a^2} + 2ab + {b^2}}}} \right):\left( {\dfrac{a}{{a + b}} – \dfrac{{{a^2}}}{{{a^2} – {b^2}}}} \right)$ есептеңіз.
Шешуі: $$ = \left( {\dfrac{{{a^2}}}{{a + b}} – \dfrac{{{a^3}}}{{{{(a + b)}^2}}}} \right):\left( {\dfrac{a}{{a + b}} – \dfrac{{{a^2}}}{{(a + b)(a – b)}}} \right) = $$ $$ = \dfrac{{{a^3} + {a^2}b – {a^3}}}{{{{(a + b)}^2}}}:\dfrac{{{a^2} – ab – {a^2}}}{{(a + b)(a – b)}} = $$ $$ = \dfrac{{{a^2}b}}{{{{(a + b)}^2}}} \cdot \dfrac{{(a + b)(a – b)}}{{ – ab}} = \dfrac{{a(a – b)}}{{ – (a + b)}} = \dfrac{{a(b – a)}}{{a + b}} = $$ $$ = \left\| {\begin{array}{*{20}{l}}{a = – 2,5}\\{b = 0,5}\end{array}} \right\| = \dfrac{{ – 2,5(0,5 + 2,5)}}{{ – 2,5 + 0,5}} = \dfrac{{ – 2,5 \cdot 3}}{{ – 2}} = 3,75$$
№ 29 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{49}}{{{a^3} + 27}} – \dfrac{{a + 3}}{{{a^2} + 9 – 3a}}} \right) \cdot \dfrac{{a\left( {{a^3} + 27} \right)}}{ {16 – {a^2}}} + \dfrac{{4 – 9a – {a^2}}}{{a + 4}}}$
Шешуі: $${ = \left( {\dfrac{{49}}{{(a + 3)\left( {{a^2} – 3a + 9} \right)}} – \dfrac{{a + 3}}{{{a^2} – 3a + 9}}} \right) \cdot \dfrac{{a\left( {{a^3} + 27} \right)}}{{16 – {a^2}}} + \dfrac{{4 – 9a – {a^2}}}{{a + 4}} = }$$ $${ = \dfrac{{49 – {{(a + 3)}^2}}}{{(a + 3)\left( {{a^2} – 3a + 9} \right)}} \cdot \dfrac{{a(a + 3)\left( {{a^2} – 3a + 9} \right)}}{{(4 – a)(4 + a)}} + \dfrac{{4 – 9a – {a^2}}}{{a + 4}} = }$$ $${ = \dfrac{{(7 – a – 3)(7 + a + 3) \cdot a}}{{(4 – a)(a + 4)}} + \dfrac{{4 – 9a – {a^2}}}{{a + 4}} = }$$ $${ = \dfrac{{(4 – a)(10 + a) \cdot a}}{{(4 – a)(a + 4)}} + \dfrac{{4 – 9a – {a^2}}}{{a + 4}} = }$$ $${ = \dfrac{{a(10 + a)}}{{a + 4}} + \dfrac{{4 – 9a – {a^2}}}{{a + 4}} = \dfrac{{10a + {a^2} + 4 – 9a – {a^2}}}{{a + 4}} = \dfrac{{a + 4}}{{a + 4}} = 1}$$
№ 30 Өрнекті ықшамдаңыз: $(x + 1) \cdot \left( {\dfrac{1}{{x + 1}} + \dfrac{4}{{{x^2} – 4x}} – \dfrac{5}{{{x^2} – 3x – 4}}} \right):\left( {1 – \dfrac{1}{x}} \right)$
Шешуі: $$ = (x + 1) \cdot \left( {\dfrac{1}{{x + 1}} + \dfrac{4}{{x(x – 4)}} – \dfrac{5}{{(x + 1)(x – 4)}}} \right):\dfrac{{x – 1}}{x} = $$ $$ = (x + 1) \cdot \dfrac{{x(x – 4) + 4(x + 1) – 5x}}{{x(x + 1)(x – 4)}} \cdot \dfrac{x}{{x – 1}} = $$ $$ = \dfrac{{x + 1}}{1} \cdot \dfrac{{{x^2} – 4x + 4x + 4 – 5x}}{{x(x + 1)(x – 4)}} \cdot \dfrac{x}{{x – 1}} = $$ $$ = \dfrac{{{x^2} – 5x + 4}}{{x(x – 4)}} \cdot \dfrac{x}{{x – 1}} = \dfrac{{(x – 1)(x – 4)}}{{x(x – 4)}} \cdot \dfrac{x}{{x – 1}} = 1$$
№ 31 Өрнекті ықшамдаңыз: ${\left( {\dfrac{2}{a} – \dfrac{a}{4} – \dfrac{3}{{2a}} + \dfrac{1}{{2a}}} \right):\dfrac{{2 – a}}{4} – \left( {1 + \dfrac{2}{a}} \right)}$
Шешуі: $${ = \dfrac{{8 – {a^2} – 6 + 2}}{{4a}} \cdot \dfrac{4}{{2 – a}} – \dfrac{{a + 2}}{a} = }$$ $$ = \dfrac{{4 – {a^2}}}{{4a}} \cdot \dfrac{4}{{2 – a}} – \dfrac{{a + 2}}{a} = $$ $$ = \dfrac{{(2 – a)(2 + a)}}{{4a}} \cdot \dfrac{4}{{2 – a}} – \dfrac{{a + 2}}{a} = \dfrac{{a + 2}}{a} – \dfrac{{a + 2}}{a} = 0$$
№ 32 Өрнекті ықшамдаңыз: ${\left( {\dfrac{8}{{2{a^2} – 8a}} – \dfrac{{3a + 32}}{{{a^3} – 64}}} \right):\dfrac{{a – 8}}{{{a^3} + 4{a^2} + 16a}} – \dfrac{4}{{4 – a}}}$
Шешуі: $${ = \left( {\dfrac{8}{{2a(a – 4)}} – \dfrac{{3a + 32}}{{(a – 4)\left( {{a^2} + 4a + 16} \right)}}} \right) \cdot \dfrac{{{a^3} + 4{a^2} + 16a}}{{a – 8}} + \dfrac{4}{{a – 4}} = }$$ $${ = \left( {\dfrac{4}{{a(a – 4)}} – \dfrac{{3a + 32}}{{(a – 4)({a^2} + 4a + 16)}}} \right) \cdot \dfrac{{a\left( {{a^2} + 4a + 16} \right)}}{{a – 8}} + \dfrac{4}{{a – 4}} = }$$ $${ = \dfrac{{4\left( {{a^2} + 4a + 16} \right) – a(3a + 32)}}{{a(a – 4)\left( {{a^2} + 4a + 16} \right)}} \cdot \dfrac{{a\left( {{a^2} + 4a + 16} \right)}}{{a – 8}} + \dfrac{4}{a-4}=}$$ $${ = \dfrac{{4{a^2} + 16a + 64 – 3{a^2} – 32a}}{{a – 4}} \cdot \dfrac{1}{{a – 8}} + \dfrac{4}{{a – 4}} = }$$ $${ = \dfrac{{{a^2} – 16a + 64}}{{a – 4}} \cdot \dfrac{1}{{a – 8}} + \dfrac{4}{{a – 4}} = \dfrac{{{{(a – 8)}^2}}}{{a – 4}} \cdot \dfrac{1}{{a – 8}} + \dfrac{4}{{a – 4}} = }$$ $${ = \dfrac{{a – 8}}{{a – 4}} + \dfrac{4}{{a – 4}} = \dfrac{{a – 8 + 4}}{{a – 4}} = \dfrac{{a – 4}}{{a – 4}} = 1}$$
№ 33 Өрнекті ықшамдаңыз: $\dfrac{{4{a^2} – 1}}{{{a^2} – 1}}:\left( {\dfrac{a}{{{a^2} – 2a + 1}} + \dfrac{a}{{{a^2} – 1}} – \dfrac{2}{{a + 1}}} \right)$
Шешуі: $$ = \dfrac{{(2a – 1)(2a + 1)}}{{(a – 1)(a + 1)}}:\left( {\dfrac{a}{{{{(a – 1)}^2}}} + \dfrac{a}{{(a – 1)(a + 1)}} – \dfrac{2}{{a + 1}}} \right) = $$ $$ = \dfrac{{(2a – 1)(2a + 1)}}{{(a – 1)(a + 1)}}:\dfrac{{a(a + 1) + a(a – 1) – 2{{(a – 1)}^2}}}{{{{(a – 1)}^2}(a + 1)}} = $$ $$ = \dfrac{{(2a – 1)(2a + 1)}}{{(a – 1)(a + 1)}}:\dfrac{{{a^2} + a + {a^2} – a – 2{{(a – 1)}^2}}}{{{{(a – 1)}^2}(a + 1)}} = $$ $$ = \dfrac{{(2a – 1)(2a + 1)}}{{(a – 1)(a + 1)}} \cdot \dfrac{{{{(a – 1)}^2}(a + 1)}}{{2{a^2} – 2{{(a – 1)}^2}}} = $$ $$ = (2a – 1)(2a + 1) \cdot \dfrac{{a – 1}}{{2\left( {{a^2} – {{(a – 1)}^2}} \right)}} = $$ $$ = (2a – 1)(2a + 1) \cdot \dfrac{{a – 1}}{{2(a – a + 1)(a + a – 1)}} = $$ $$ = \left( {2a – 1} \right)\left( {2a + 1} \right) \cdot \frac{{a – 1}}{{2\left( {2a – 1} \right)}} = \left( {2a + 1} \right) \cdot \frac{{a – 1}}{2} = $$ $$ = \frac{{\left( {2a + 1} \right)\left( {a – 1} \right)}}{2} = \frac{{2{a^2} – 2a + a – 1}}{2} = \frac{{2{a^2} – a – 1}}{2}$$
№ 34 Өрнекті ықшамдаңыз: $\dfrac{{{{\left( {{x^2} + 5} \right)}^2} + 4\left( {{x^2} + 5} \right) + 4}}{{{x^2} – 10x + 21}} \cdot \dfrac{{2{x^2} – 18}}{{{{\left( {{x^2} + 7} \right)}^2}}}$
Шешуі: $$ = \dfrac{{{x^4} + 10{x^2} + 25 + 4{x^2} + 20 + 4}}{{(x – 3)(x – 7)}} \cdot \dfrac{{2\left( {{x^2} – 9} \right)}}{{{{\left( {{x^2} + 7} \right)}^2}}} = $$ $$ = \dfrac{{{x^4} + 14{x^2} + 49}}{{(x – 3)(x – 7)}} \cdot \dfrac{{2(x – 3)(x + 3)}}{{{{\left( {{x^2} + 7} \right)}^2}}} = $$ $$ = \dfrac{{{{\left( {{x^2} + 7} \right)}^2}}}{{(x – 3)(x – 7)}} \cdot \dfrac{{2(x – 3)(x + 3)}}{{{{\left( {{x^2} + 7} \right)}^2}}} = \dfrac{{2(x + 3)}}{{x – 7}} = \dfrac{{2x + 6}}{{x – 7}}$$
№ 35 Өрнекті ықшамдаңыз: $\dfrac{{{y^2} + 18y + 77}}{{{y^2} – 49}} \cdot \dfrac{{5y – 35}}{{{{(y + 8)}^2} + 6(y + 8) + 9}}$
Шешуі: $$ = \dfrac{{{y^2} + 18y + 77}}{{(y – 7)(y + 7)}} \cdot \dfrac{{5(y – 7)}}{{((y + 8) + 3)^2}} = \dfrac{{{y^2} + 18y + 77}}{{y + 7}} \cdot \dfrac{5}{{{{(y + 11)}^2}}} = $$ $$ = \dfrac{{(y + 7)(y + 11)}}{{y + 7}} \cdot \dfrac{5}{{{{(y + 11)}^2}}} = \dfrac{5}{{y + 11}}$$
№ 36 Өрнекті ықшамдаңыз: $\left( {\dfrac{1}{{{{(2 – a)}^2}}} – \dfrac{2}{{{a^2} – 4}} + \dfrac{1}{{{{(2 + a)}^2}}}} \right) \cdot {\left( {{a^2} – 4} \right)^2}$
Шешуі: $$ = \left( {\dfrac{1}{{{{(a – 2)}^2}}} – \dfrac{2}{{(a – 2)(a + 2)}} + \dfrac{1}{{{{(a + 2)}^2}}}} \right) \cdot \left((a – 2)(a + 2)\right)^2 = $$ $$ = \dfrac{{{{(a + 2)}^2} – 2(a – 2)(a + 2) + {{(a – 2)}^2}}}{{{{(a – 2)}^2}{{(a + 2)}^2}}} \cdot {(a – 2)^2} \cdot {(a + 2)^2} = $$ $$ = {a^2} + 4a + 4 – 2{a^2} + 8 + {a^2} – 4a + 4 = 16$$
№ 37 Өрнекті ықшамдаңыз: ${\dfrac{1}{{(a – b)(a – c)}} + \dfrac{1}{{(b – c)(b – a)}} + \dfrac{1}{{(c – a)(c – b)}}}$
Шешуі: $${ = \dfrac{1}{{(a – b)(a – c)}} – \dfrac{1}{{(a – b)(b – c)}} + \dfrac{1}{{(a – c)(b – c)}} = }$$ $${ = \dfrac{{(b – c) – (a – c) + (a – b)}}{{(a – b)(a – c)(b – c)}} = \dfrac{{b – c – a + c + a – b}}{{(a – b)(a – c)(b – c)}} = }$$ $${ = \dfrac{0}{{(a – b)(a – c)(b – c)}} = 0}$$
№ 38 Өрнекті ықшамдаңыз: $\left( {{x^2} – {y^2} – {z^2} + 2yz} \right):\dfrac{{x + y – z}}{{x + y + z}}$
Шешуі: $$ = \left( {{x^2} – \left( {{y^2} – 2yz + {z^2}} \right)} \right):\dfrac{{x + y – z}}{{x + y + z}} = $$ $${ = \left( {{x^2} – {{(y – z)}^2}} \right) \cdot \dfrac{{x + y + z}}{{x + y – z}} = }$$ $$ = (x – y + z)(x + y – z) \cdot \dfrac{{x + y + z}}{{x + y – z}} = $$ $$ = ((x + z) – y) \cdot ((x + z) + y) = {(x + z)^2} – {y^2}$$
№ 39 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{2x + 1}}{{x + 2}} – \dfrac{{4x + 2}}{{4 – {x^2}}}} \right):\dfrac{{2x + 1}}{{x – 2}} + \dfrac{2}{{x + 2}}}$
Шешуі: $${ = \left( {\dfrac{{2x + 1}}{{x + 2}} + \dfrac{{4x + 2}}{{{x^2} – 4}}} \right) \cdot \dfrac{{x – 2}}{{2x + 1}} + \dfrac{2}{{x + 2}} = }$$ $${ = \left( {\dfrac{{2x + 1}}{{x + 2}} + \dfrac{{4x + 2}}{{(x – 2)(x + 2)}}} \right) \cdot \dfrac{{x – 2}}{{2x + 1}} + \dfrac{2}{{x + 2}} = }$$ $${ = \dfrac{{(2x + 1)(x – 2) + 2(2x + 1)}}{{(x – 2)(x + 2)}} \cdot \dfrac{{x – 2}}{{2x + 1}} + \dfrac{2}{{x + 2}} = }$$ $$ = \dfrac{{(2x + 1)(x – 2 + 2)}}{{x + 2}} \cdot \dfrac{1}{{2x + 1}} + \dfrac{2}{{x + 2}} = $$ $$ = \dfrac{x}{{x + 2}} + \dfrac{2}{{x + 2}} = \dfrac{{x + 2}}{{x + 2}} = 1$$
№ 40 Өрнекті ықшамдаңыз: $\left( {\dfrac{9}{{{x^2} – 9}} + \dfrac{3}{{{{(3 – x)}^2}}}} \right):\dfrac{6}{{{{(x – 3)}^2}}} + \dfrac{{1 – 2x}}{{3 + x}}$
Шешуі: $$ = \left( {\dfrac{9}{{(x – 3)(x + 3)}} + \dfrac{3}{{{{(x – 3)}^2}}}} \right) \cdot \dfrac{{{{(x – 3)}^2}}}{6} + \dfrac{{1 – 2x}}{{3 + x}} = $$ $$ = \dfrac{{9(x – 3) + 3(x + 3)}}{{{{(x – 3)}^2}(x + 3)}} \cdot \dfrac{{{{(x – 3)}^2}}}{6} + \dfrac{{1 – 2x}}{{3 + x}} = $$ $$ = \dfrac{{9x – 27 + 3x + 9}}{{x + 3}} \cdot \dfrac{1}{6} + \dfrac{{1 – 2x}}{{3 + x}} = $$ $$ = \dfrac{{12x – 18}}{{x + 3}} \cdot \dfrac{1}{6} + \dfrac{{1 – 2x}}{{3 + x}} = \dfrac{{6(2x – 3)}}{{x + 3}} \cdot \dfrac{1}{6} + \dfrac{{1 – 2x}}{{3 + x}} = $$ $$ = \dfrac{{2x – 3}}{{x + 3}} + \dfrac{{1 – 2x}}{{x + 3}} = \dfrac{{2x – 3 + 1 – 2x}}{{x + 3}} = – \dfrac{2}{{x + 3}}$$
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Есеп шешімдерінің авторы Сейдегалымова Жанар Назарбековна