Рационал алгебралық өрнектерді түрлендіру (Рустюмова 1.2.5)

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Құрамында лері бар өрнектерді ықшамдаңыз.

№ 1 Өрнекті ықшамдаңыз: ${(a + b)^{ - 1}} \cdot \dfrac{{{a^{ - 2}} + {b^{ - 2}}}}{{{a^{ - 1}} + {b^{ - 1}}}}:{\left( {\dfrac{{ab}}{{{a^2} + {b^2}}}} \right)^{ - 1}} \cdot {\left( {\dfrac{{2ab}}{{a + b}}} \right)^{ - 1}}$

Шешуі: $$1)\quad \frac{{{a^{ - 2}} + {b^{ - 2}}}}{{{a^{ - 1}} + {b^{ - 1}}}} = \frac{{\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}}{{\frac{1}{a} + \frac{1}{b}}} = \frac{{\frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}}}}{{\frac{{a + b}}{{ab}}}} = \frac{{ab \cdot \left( {{a^2} + {b^2}} \right)}}{{{{(ab)}^2} \cdot (a + b)}} = \frac{{{a^2} + {b^2}}}{{ab(a + b)}}$$ $$2)\quad {(a + b)^{ - 1}} \cdot \frac{{{a^2} + {b^2}}}{{ab(a + b)}} = \frac{1}{{a + b}} \cdot \frac{{{a^2} + {b^2}}}{{ab(a + b)}} = \frac{{{a^2} + {b^2}}}{{ab{{(a + b)}^2}}}$$ $$3)\quad \frac{{{a^2} + {b^2}}}{{ab{{(a + b)}^2}}}:{\left( {\frac{{ab}}{{{a^2} + {b^2}}}} \right)^{ - 1}} = \frac{{{a^2} + {b^2}}}{{ab{{(a + b)}^2}}} \cdot \frac{{ab}}{{{a^2} + {b^2}}} = \frac{1}{{{{(a + b)}^2}}}$$ $$4)\quad \frac{1}{{{{(a + b)}^2}}} \cdot {\left( {\frac{{2ab}}{{a + b}}} \right)^{ - 1}} = \frac{1}{{{{(a + b)}^2}}} \cdot \frac{{a + b}}{{2ab}} = \frac{1}{{2(a + b)ab}}$$ Жауабы: $\dfrac{1}{{2(a + b)ab}}$

№ 2 Өрнекті ықшамдаңыз: $\left( {{a^{ - 2}} - {b^{ - 2}}} \right) \cdot {\left( {{b^{ - 1}} - {a^{ - 1}}} \right)^{ - 1}}$

Шешуі: $$ = \left( {\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}}} \right) \cdot {\left( {\frac{1}{b} - \frac{1}{a}} \right)^{ - 1}} = \frac{{{b^2} - {a^2}}}{{{a^2}{b^2}}} \cdot {\left( {\frac{{a - b}}{{ab}}} \right)^{ - 1}} = $$ $$ = \frac{{(b - a)(b + a)}}{{{{(ab)}^2}}} \cdot \frac{{ab}}{{a - b}} = - \frac{{b + a}}{{ab}}$$ Жауабы: $ - \dfrac{{b + a}}{{ab}}$

№ 3 Өрнекті ықшамдаңыз: $\dfrac{{{x^{ - 2}} - {y^{ - 2}}}}{{{x^{ - 1}} + {y^{ - 1}}}}:\dfrac{{{x^{ - 1}} \cdot {y^{ - 1}}}}{{{{(y - x)}^{ - 1}}}}$

Шешуі: $$ = \frac{{\frac{1}{{{x^2}}} - \frac{1}{{{y^2}}}}}{{\frac{1}{x} + \frac{1}{y}}} \cdot \frac{{{{(y - x)}^{ - 1}}}}{{{x^{ - 1}} \cdot {y^{ - 1}}}} = \frac{{\frac{{{y^2} - {x^2}}}{{{{(xy)}^2}}}}}{{\frac{{y + x}}{{xy}}}} \cdot \frac{{\frac{1}{{y - x}}}}{{\frac{1}{{xy}}}} = $$ $$ = \frac{{xy(y - x)(y + x)}}{{{{(xy)}^2} \cdot (x + y)}} \cdot \frac{{xy}}{{y - x}} = 1.$$ Жауабы: $1$

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

Шешуі: $$ = \frac{{\frac{{{a^3}}}{b} - \frac{{{b^3}}}{a}}}{{\frac{a}{b} + \frac{b}{a}}} \cdot \frac{{ab}}{{{a^2} - {b^2}}} = \frac{{\frac{{{a^4} - {b^4}}}{{ab}}}}{{\frac{{{a^2} + {b^2}}}{{ab}}}} \cdot \frac{{ab}}{{{a^2} - {b^2}}} = $$ $$ = \frac{{ab\left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)}}{{ab\left( {{a^2} + {b^2}} \right)}} \cdot \frac{{ab}}{{{a^2} - {b^2}}} = ab.$$ Жауабы: $ab$

№ 5 Өрнекті ықшамдаңыз: $\dfrac{{{a^{ - 2}}{b^3} - a}}{{\frac{1}{{{a^3}{b^{ - 2}}}} + {a^{ - 2}}b + {a^{ - 1}}}}$

Шешуі: $$ = \frac{{\frac{{{b^3}}}{{{a^2}}} - a}}{{\frac{{{b^2}}}{{{a^3}}} + \frac{b}{{{a^2}}} + \frac{1}{a}}} = \frac{{\frac{{{b^3} - {a^3}}}{{{a^2}}}}}{{\frac{{{b^2} + ab + {a^2}}}{{{a^3}}}}} = $$ $${ = \frac{{{a^3}\left( {{b^3} - {a^3}} \right)}}{{{a^2}\left( {{b^2} + ab + {a^2}} \right)}} = \frac{{a(b - a)\left( {{b^2} + ab + {a^2}} \right)}}{{{b^2} + ab + {a^2}}}}$$ $${ = a(b - a){\rm{. }}}$$ Жауабы: ${a(b - a)}$

№ 6 Өрнекті ықшамдаңыз: ${\left( {\dfrac{{{a^{ - 1}} - {b^{ - 1}}}}{{{a^{ - 3}} + {b^{ - 3}}}}:\dfrac{{{a^2}{b^2}}}{{{{(a + b)}^2} - 3ab}} + 1} \right)^{ - 1}} - 0,5$

Шешуі: $$1)\quad \frac{{{a^{ - 1}} - {b^{ - 1}}}}{{{a^{ - 3}} + {b^{ - 3}}}} = \frac{{\frac{1}{a} + \frac{1}{b}}}{{\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}}}} = \frac{{\frac{{a + b}}{{ab}}}}{{\frac{{{a^3} + {b^3}}}{{{{(ab)}^3}}}}} = $$ $$ = \frac{{{{(ab)}^3}(a + b)}}{{ab\left( {{a^3} + {b^3}} \right)}} = \frac{{{{(ab)}^2}(a + b)}}{{(a + b)\left( {{a^2} - ab + {b^2}} \right)}} = \frac{{{{(ab)}^2}}}{{{a^2} - ab + {b^2}}}$$ $$2)\quad \frac{{{{(ab)}^2}}}{{{a^2} - ab + {b^2}}}:\frac{{{a^2}{b^2}}}{{{{(a + b)}^2} - 3ab}} = \frac{{{a^2}{b^2}}}{{{a^2} - ab + {b^2}}} \cdot \frac{{{{(a + b)}^2} - 3ab}}{{{a^2}{b^2}}} = $$ $$ = \frac{{{a^2} + 2ab + {b^2} - 3ab}}{{{a^2} - ab + {b^2}}} = \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} = 1$$ $$3)\quad {(1 + 1)^{ - 1}} = {2^{ - 1}} = \frac{1}{2}$$ $$4)\quad \frac{1}{2} - 0,5 = 0,5 - 0,5 = 0.$$ Жауабы: $0$

№ 7 Егер $m=0,003; \quad n=0,007$ болса, $\dfrac{{{m^{ - 2}}{n^{ - 1}} - {m^{ - 1}}{n^{ - 2}}}}{{{m^{ - 2}} - {n^{ - 2}}}} - \dfrac{1}{m}{\left( {m{n^{ - 1}} + 2 + {m^{ - 1}}n} \right)^{ - 1}}$ мәнін табыңыз.

Шешуі: $$1)\quad \frac{{{m^{ - 2}}{n^{ - 1}} - {m^{ - 1}}{n^{ - 2}}}}{{{m^{ - 2}} - {n^{ - 2}}}} = \frac{{\frac{1}{{{m^2}n}} - \frac{1}{{m{n^2}}}}}{{\frac{1}{{{m^2}}} - \frac{1}{{{n^2}}}}} = \frac{{\frac{{n - m}}{{{m^2}{n^2}}}}}{{\frac{{{n^2} - {m^2}}}{{{m^2}{n^2}}}}} = $$ $$ = \frac{{{m^2}{n^2}(n - m)}}{{{m^2}{n^2}(n - m)(n + m)}} = \frac{1}{{n + m}}$$ $$2)\quad {\left( {m{n^{ - 1}} + 2 + {m^{ - 1}}n} \right)^{ - 1}} = {\left( {\frac{m}{n} + 2 + \frac{n}{m}} \right)^{ - 1}} = $$ $$ = {\left( {\frac{{{m^2} + 2mn + {n^2}}}{{mn}}} \right)^{ - 1}} = {\left( {\frac{{{{(m + n)}^2}}}{{mn}}} \right)^{ - 1}} = \frac{{mn}}{{{{(m + n)}^2}}}$$ $$3)\quad \frac{1}{m} \cdot \frac{{mn}}{{{{(m + n)}^2}}} = \frac{n}{{{{(m + n)}^2}}}$$ $$4)\quad \frac{1}{{n + m}} - \frac{n}{{{{(m + n)}^2}}} = \frac{{m + n - n}}{{{{(m + n)}^2}}} = \frac{m}{{{{(m + n)}^2}}}$$ $$5)\quad \frac{{0,003}}{{{{(0,003 + 0,007)}^2}}} = \frac{{0,003}}{{0,0001}} = \frac{{0,003 \cdot 10000}}{{0,0001 \cdot 10000}} = \frac{{30}}{1} = 30.$$ Жауабы: $30$

№ 8 Өрнекті ықшамдаңыз: $\left( {\dfrac{{{a^{ - 1}} - {b^{ - 1}}}}{{{a^{ - 1}} + {b^{ - 1}}}} - \dfrac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 1}} - {b^{ - 1}}}}} \right) \cdot {\left( {\dfrac{{4ab}}{{{b^2} - {a^2}}}} \right)^{ - 1}}$

Шешуі: $$1)\quad \frac{{{a^{ - 1}} - {b^{ - 1}}}}{{{a^{ - 1}} + {b^{ - 1}}}} = \frac{{\frac{1}{a} - \frac{1}{b}}}{{\frac{1}{a} + \frac{1}{b}}} = \frac{{\frac{{b - a}}{{ab}}}}{{\frac{{b + a}}{{ab}}}} = \frac{{ab(b - a)}}{{ab(b + a)}} = \frac{{b - a}}{{a + b}}$$ $$2)\quad \frac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 1}} - {b^{ - 1}}}} = \frac{{\frac{1}{a} + \frac{1}{b}}}{{\frac{1}{a} - \frac{1}{b}}} = \frac{{\frac{{b + a}}{{ab}}}}{{\frac{{b - a}}{{ab}}}} = \frac{{ab(b + a)}}{{ab(b - a)}} = \frac{{a + b}}{{b - a}}$$ $$3)\quad \frac{{b - a}}{{a + b}} - \frac{{a + b}}{{b - a}} = \frac{{{{(b - a)}^2} - {{(a + b)}^2}}}{{(a + b)(b - a)}} = \frac{{{b^2} - 2ab + {a^2} - \left( {{a^2} + 2ab + {b^2}} \right)}}{{{b^2} - {a^2}}} = $$ $$ = \frac{{{b^2} - 2ab + {a^2} - {a^2} - 2ab - {b^2}}}{{{b^2} - {a^2}}} = \frac{{ - 4ab}}{{{b^2} - {a^2}}}$$ $$4)\quad \frac{{ - 4ab}}{{{b^2} - {a^2}}} \cdot {\left( {\frac{{4ab}}{{{b^2} - {a^2}}}} \right)^{ - 1}} = - \frac{{4ab}}{{{b^2} - {a^2}}} \cdot \frac{{{b^2} - {a^2}}}{{4ab}} = - 1.$$ Жауабы: $-1$

№ 9 Өрнекті ықшамдаңыз: $\left( {{x^2} - {a^{ - 1}}x + {a^{ - 2}}} \right) \cdot \left( {{x^{ - 1}} + a} \right) - x \cdot {(ax)^{ - 2}}$

Шешуі: $$ = \left( {{x^2} - \frac{x}{a} + \frac{1}{{{a^2}}}} \right) \cdot \left( {\frac{1}{x} + a} \right) - \frac{x}{{{a^2}{x^2}}} = $$ $$ = \frac{{{a^2}{x^2} - ax + 1}}{{{a^2}}} \cdot \frac{{1 + ax}}{x} - \frac{1}{{{a^2}x}} = $$ $$ = \frac{{{a^2}{x^2} - ax + 1 + {a^3}{x^3} - {a^2}{x^2} + ax}}{{{a^2}x}} - \frac{1}{{{a^2}x}} = $$ $$ = \frac{{1 + {a^3}{x^3} - 1}}{{{a^2}x}} = a{x^2}$$ Жауабы: $ax^2$

№ 10 Өрнекті ықшамдаңыз: ${\dfrac{{{a^{ - 1}} - {x^{ - 1}}}}{{{a^{ - 3}} + {x^{ - 3}}}}:{{\left( {\dfrac{{x{a^{ - 2}} + a{x^{ - 2}}}}{{x - a}}} \right)}^{ - 1}}}$

Шешуі: $${ = \frac{{\frac{1}{a} - \frac{1}{x}}}{{\frac{1}{{{a^3}}} + \frac{1}{{{x^3}}}}} \cdot \frac{{\frac{x}{{{a^2}}} + \frac{a}{{{x^2}}}}}{{x - a}} = \frac{{\frac{{x - a}}{{ax}}}}{{\frac{{{x^3} + {a^3}}}{{{a^3}{x^3}}}}} \cdot \frac{{\frac{{{x^3} + {a^3}}}{{{a^2}{x^2}}}}}{{x - a}} = }$$ $${ = \frac{{{a^3}{x^3}(x - a)}}{{ax\left( {{x^3} + {a^3}} \right)}} \cdot \frac{{{x^3} + {a^3}}}{{{a^2}{x^2}(x - a)}} = 1}$$ Жауабы: $1$

№ 11 Егер $x = 0,24;\quad y = \dfrac{5}{{12}}$ болса, $\dfrac{{\left( {x{y^{ - 1}} + y{x^{ - 1}} + 1} \right) \cdot {{\left( {{x^{ - 1}} - {y^{ - 1}}} \right)}^2}}}{{{x^2}{y^{ - 2}} + {y^2}{x^{ - 2}} - x{y^{ - 1}} - y{x^{ - 1}}}}$ еептеңіз.

Шешуі: $$1)x{y^{ - 1}} + y{x^{ - 1}} + 1 = \frac{x}{y} + \frac{y}{x} + 1 = \frac{{{x^2} + {y^2} + xy}}{{xy}}$$ $$2){\left( {{x^{ - 1}} - {y^{ - 1}}} \right)^2} = {\left( {\frac{1}{x} - \frac{1}{y}} \right)^2} = \frac{{{{(y - x)}^2}}}{{{x^2}{y^2}}}$$ $$3)\frac{{{x^2} + {y^2} + xy}}{{xy}} \cdot \frac{{{{(y - x)}^2}}}{{{x^2}{y^2}}} = \frac{{\left( {{x^2} + xy + {y^2}} \right){{(x - y)}^2}}}{{{x^3}{y^3}}}$$ $$4){x^2}{y^{ - 2}} + {y^2}{x^{ - 2}} - x{y^{ - 1}} - y{x^{ - 1}} = \frac{{{x^2}}}{{{y^2}}} + \frac{{{y^2}}}{{{x^2}}} - \frac{x}{y} - \frac{y}{x} = $$ $${ = \frac{{x(x - y)}}{{{y^2}}} + \frac{{y(y - x)}}{{{x^2}}} = \frac{{x(x - y)}}{{{y^2}}} - \frac{{y(x - y)}}{{{x^2}}} = }$$ $${ = \frac{{{x^3}(x - y) - {y^3}(x - y)}}{{{x^2}{y^2}}} = \frac{{(x - y)\left( {{x^3} - {y^3}} \right)}}{{{x^2}{y^2}}}}$$ $${5)\quad \frac{{\frac{{\left( {{x^2} + xy + {y^2}} \right){{(x - y)}^2}}}{{{x^3}{y^3}}}}}{{\frac{{(x - y)\left( {{x^3} - {y^3}} \right)}}{{{x^2}{y^2}}}}} = \frac{{{x^2}{y^2}\left( {{x^2} + xy + {y^2}} \right){{(x - y)}^2}}}{{(x - y)\left( {{x^3} - {y^3}} \right){x^3}{y^3}}} = }$$ $${ = \frac{{(x - y)\left( {{x^2} + xy + {y^2}} \right)}}{{xy(x - y)\left( {{x^2} + xy + {y^3}} \right)}} = \frac{1}{{xy}}}$$ $${6)\quad \frac{1}{{0,24 \cdot \frac{5}{{12}}}} = \frac{1}{{\frac{{24}}{{100}} \cdot \frac{5}{{12}}}} = \frac{1}{{\frac{2}{{20}}}} = \frac{{20}}{2} = 10}$$ Жауабы: $10$

№ 12 Өрнекті ықшамдаңыз: $\dfrac{{{{\left( {a{b^{ - 3}} - {a^{ - 3}}b} \right)}^{ - 1}} \cdot \left( {{a^{ - 2}} + {b^{ - 2}}} \right)}}{{{{\left( {{b^{ - 2}} - {a^{ - 2}}} \right)}^{ - 1}}}}$

Шешуі: $$1)\quad {\left( {a{b^{ - 3}} - {a^{ - 3}}b} \right)^{ - 1}} = {\left( {\frac{a}{{{b^3}}} - \frac{b}{{{a^3}}}} \right)^{ - 1}} = {\left( {\frac{{{a^4} - {b^4}}}{{{a^3}{b^3}}}} \right)^{ - 1}} = \frac{{{a^3}{b^3}}}{{{a^4} - {b^4}}}$$ $$2)\quad {a^{ - 2}} + {b^{ - 2}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{{{b^2} + {a^2}}}{{{a^2}{b^2}}}$$ $$3)\quad \frac{{{a^3}{b^3}}}{{{a^4} - {b^4}}} \cdot \frac{{{b^2} + {a^2}}}{{{a^2}{b^2}}} = \frac{{ab\left( {{a^2} + {b^2}} \right)}}{{\left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)}} = \frac{{ab}}{{{a^2} - {b^2}}}$$ $$4)\quad {\left( {{b^{ - 2}} - {a^{ - 2}}} \right)^{ - 1}} = {\left( {\frac{1}{{{b^2}}} - \frac{1}{{{a^2}}}} \right)^{ - 1}} = {\left( {\frac{{{a^2} - {b^2}}}{{{a^2}{b^2}}}} \right)^{ - 1}} = \frac{{{a^2}{b^2}}}{{{a^2} - {b^2}}}$$ $$5)\quad \frac{{\frac{{ab}}{{{a^2} - {b^2}}}}}{{\frac{{{a^2}{b^2}}}{{{a^2} - {b^2}}}}} = \frac{{ab\left( {{a^2} - {b^2}} \right)}}{{{a^2}{b^2}\left( {{a^2} - {b^2}} \right)}} = \frac{1}{{ab}}$$ Жауабы: $\dfrac{1}{{ab}}$

№ 13 Егер $a=10, \quad b=2$ болса, ${\dfrac{{{{\left( {{a^{ - 1}}{b^2} + {a^3}{b^{ - 4}}} \right)}^2}}}{{{b^3}{a^{ - 4}} + 2{b^{ - 3}} + {a^4}{b^{ - 9}}}}}$ есептеңіз.

Шешуі: $${\frac{{{{\left( {{a^{ - 1}}{b^2} + {a^3}{b^{ - 4}}} \right)}^2}}}{{{b^3}{a^{ - 4}} + 2{b^{ - 3}} + {a^4}{b^{ - 9}}}} = \frac{{{{\left( {\frac{{{b^2}}}{a} + \frac{{{a^3}}}{{{b^4}}}} \right)}^2}}}{{\frac{{{b^3}}}{{{a^4}}} + \frac{2}{{{b^3}}} + \frac{{{a^4}}}{{{b^9}}}}} = }$$ $${ = \frac{{\frac{{{{\left( {{b^6} + {a^4}} \right)}^2}}}{{{a^2}{b^8}}}}}{{\frac{{{b^{12}} + 2{a^4}{b^6} + {a^8}}}{{{a^4}{b^9}}}}} = \frac{{{a^4}{b^9}{{\left( {{b^6} + {a^4}} \right)}^2}}}{{{a^2}{b^8}\left( {{b^{12}} + 2{a^4}{b^6} + {a^8}} \right)}} = }$$ $${ = \frac{{{a^2}b{{\left( {{b^6} + {a^4}} \right)}^2}}}{{{{\left( {{b^6} + {a^4}} \right)}^2}}} = {a^2}b}$$ $${{{( - 10)}^2} \cdot 2 = 100 \cdot 2 = 200}$$ Жауабы: $200$

№ 14 Егер $b=-4; \quad a=\dfrac{2}{3}$ болса, ${\dfrac{{{a^{ - 1}} - 27{b^3}{a^{ - 4}}}}{{{a^{ - 1}} + 3{a^{ - 2}}b + 9{a^{ - 3}}{b^2}}}}$ есептеңіз.

Шешуі: $${1)\quad {a^{ - 1}} - 27{b^3}{a^{ - 4}} = \frac{1}{a} - \frac{{27{b^3}}}{{{a^4}}} = \frac{{{a^3} - 27{b^3}}}{{{a^4}}}}$$ $${2)\quad {a^{ - 1}} + 3{a^{ - 2}}b + 9{a^{ - 3}}{b^2} = \frac{1}{a} + \frac{{3b}}{{{a^2}}} + \frac{{9{b^2}}}{{{a^3}}} = \frac{{{a^2} + 3ab + 9{b^2}}}{{{a^3}}}}$$ $${3)\quad \frac{{\frac{{{a^3} - 27{b^3}}}{{{a^4}}}}}{{\frac{{{a^2} + 3ab + 9{b^2}}}{{{a^3}}}}} = \frac{{{a^3}(a - 3b)\left( {{a^2} + 3ab + 9{b^2}} \right)}}{{{a^4}\left( {{a^2} + 3ab + 9{b^2}} \right)}} = \frac{{a - 3b}}{a}}$$ $${4)\quad \frac{{\frac{2}{3} - 3 \cdot ( - 4)}}{{\frac{2}{3}}} = \frac{{\frac{2}{3} + 12}}{{\frac{2}{3}}} = \frac{{12\frac{2}{3}}}{{\frac{2}{3}}} = \frac{{\frac{{38}}{3}}}{{\frac{2}{3}}} = \frac{{3 \cdot 38}}{{3 \cdot 2}} = 19}$$ Жауабы: $19$

№ 15 Егер $a = b = \sqrt 2 $ болса, $\left( {2{a^{ - 3}} - {b^{ - 2}}} \right) \cdot \left( {\dfrac{{{a^{ - 3}}}}{{{2^{ - 1}}}} + \dfrac{1}{{{b^2}}}} \right) \cdot \left( {\dfrac{1}{{{b^4}}} + 4{a^{ - 6}}} \right)$ есептеңіз.

Шешуі: $$1)\quad 2{a^{ - 3}} - {b^{ - 2}} = \frac{2}{{{a^3}}} - \frac{1}{{{b^2}}} = \frac{{2{b^2} - {a^3}}}{{{a^3}{b^2}}}$$ $$2)\quad \frac{{{a^{ - 3}}}}{{{2^{ - 1}}}} + \frac{1}{{{b^2}}} = \frac{2}{{{a^3}}} + \frac{1}{{{b^2}}} = \frac{{2{b^2} + {a^3}}}{{{a^3}{b^2}}}$$ $$3)\quad \frac{1}{{{b^4}}} + 4{a^{ - 6}} = \frac{1}{{{b^4}}} + \frac{4}{{{a^6}}} = \frac{{{a^6} + 4{b^4}}}{{{a^6}{b^4}}}$$ $$4)\quad \frac{{2{b^2} - {a^3}}}{{{a^3}{b^2}}} \cdot \frac{{2{b^2} + {a^3}}}{{{a^3}{b^2}}} \cdot \frac{{{a^6} + 4{b^4}}}{{{a^6} \cdot {b^4}}} = $$ $$ = \frac{{\left( {4{b^4} - {a^6}} \right)\left( {{a^6} + 4{b^4}} \right)}}{{{a^6}{b^4} \cdot {a^6} \cdot {b^4}}} = \frac{{16{b^8} - {a^{12}}}}{{{a^{12}}{b^8}}}$$ $$5)\quad \frac{{16 \cdot {{\sqrt 2 }^8} - {{\sqrt 2 }^{12}}}}{{{{\sqrt 2 }^{12}} \cdot {{\sqrt 2 }^8}}} = \frac{{16 \cdot {2^4} - {2^6}}}{{{2^6} \cdot {2^4}}} = \frac{{{2^4} \cdot {2^4} - {2^6}}}{{{2^6} \cdot {2^4}}} = $$ $$ = \frac{{{2^8} - {2^6}}}{{{2^6} \cdot {2^4}}} = \frac{{{2^6}\left( {{2^2} - 1} \right)}}{{{2^6} \cdot {2^4}}} = \frac{{4 - 1}}{{16}} = \frac{3}{{16}}.$$ Жауабы: $\dfrac{3}{16}$

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