Задание 569
Доказываем. Докажите тождество:
- \({\largeа)}\ \frac{a^2+b^2}{ab}\cdot\left(\frac{6a+b}{a^2-b^2}:\frac{6a^3+b^3+a^2b+6ab^2}{2ab^2-2a^2b}+\frac{a+b}{a^2+b^2}\right)=\frac{a^2+b^2}{ab(a+b)};\)
- \({\largeб)}\ \left(\frac{x}{xy+y^2}-\frac{x^2+y^2}{x^3-xy^2}+\frac{y}{x^2-xy}\right):\frac{x^2-2xy+y^2}{x^3+y^3}=\frac{x^2-xy+y^2}{y(x-y)};\)
- \({\largeв)}\ \left(\frac{2x^2y+2xy^2}{7x^3+x^2y+7xy^2+y^3}\cdot\frac{7x+y}{x^2-y^2}+\frac{x-y}{x^2+y^2}\right)\cdot(x^2-y^2)=x+y;\)
- \({\largeг)}\ \left(\frac{5}{a^2-2a-ax+2x}-\frac{1}{8-8a+2a^2}\cdot\frac{20-10a}{x-2}\right):\frac{25}{x^3-8}=\frac{x^2+2x+4}{5(a-x)};\)
- \({\largeд)}\ \left(\frac{3a}{9-3x-3a+ax}-\frac{1}{a^2-9}:\frac{x-a}{3a^2+9a}\right)\cdot\frac{x^3-27}{3a}=\frac{x^2+3x+9}{a-x}.\)
Источник заимствования: Алгебра. 7 класс. Учебник для общеобразовательных организаций / С.М. Никольский, М.К. Потапов, Н.Н. Решетников, А.В. Шевкин – Просвещение, 2013. – 147 c. ISBN 978-5-09-027739-6
А+АА-
Решение:
- \({\largeа)}\ \frac{a^2+b^2}{ab}\cdot\left(\frac{6a+b}{a^2-b^2}:\frac{6a^3+b^3+a^2b+6ab^2}{2ab^2-2a^2b}+\frac{a+b}{a^2+b^2}\right)=\frac{a^2+b^2}{ab}\cdot\left(\frac{6a+b}{ (a+b)(a-b)}\cdot\frac{ 2ab(b-a)}{ 6a(a^2+b^2)+b(b^2+a^2)}+\frac{a+b}{a^2+b^2}\right)=\frac{a^2+b^2}{ab}\cdot\left(\frac{ {-}(6a+b)2ab(a-b)}{ (a+b)(a-b)(a^2+b^2)(6a+b)}+\frac{a+b}{a^2+b^2}\right)=\frac{a^2+b^2}{ab}\cdot\left(\frac{ {-}2ab^{\backslash1}}{ (a+b)(a^2+b^2)\phantom{^{\backslash1}}}+\frac{a+b^{\backslash{a\ +\ b}}}{a^2+b^2\phantom{^{\backslash{a\ +\ b}}}}\right)=\frac{a^2+b^2}{ab}\cdot\left(\frac{ {-}2ab}{ (a+b)(a^2+b^2)}+\frac{ (a+b)(a+b)}{ (a+b)(a^2+b^2)}\right)=\frac{a^2+b^2}{ab}\cdot\frac{{-}2ab+a^2+2ab+b^2}{ (a+b)(a^2+b^2)}=\frac{a^2+b^2}{ab}\cdot\frac{a^2+b^2}{ (a+b)(a^2+b^2)}=\frac{a^2+b^2}{ ab(a+b)}\)
Что и требовалось доказать.
- \({\largeб)}\ \left(\frac{x}{xy+y^2}-\frac{x^2+y^2}{x^3-xy^2}+\frac{y}{x^2-xy}\right):\frac{x^2-2xy+y^2}{x^3+y^3}=\left(\frac{x}{ y(x+y)}-\frac{x^2+y^2}{ x(x^2-y^2)}+\frac{y}{ x(x-y)}\right):\frac{ (x-y)^2}{ (x+y)(x^2-xy+y^2)}=\left(\frac{x^{\backslash{ x(x\ -\ y)}}}{ y(x+y)\phantom{^{\backslash{ x(x\ -\ y)}}}}-\frac{x^2+y^{2\backslash{y}}}{ x(x+y)(x-y)\phantom{^{\backslash{y}}}}+\frac{y^{\backslash{ y(x\ +\ y)}}}{ x(x-y)\phantom{^{\backslash{ y(x\ +\ y)}}}}\right)\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\left(\frac{ x\cdot{x}(x\ -\ y)}{ xy(x+y)(x-y)}-\frac{ (x^2+y^2)\cdot{y}}{ xy(x+y)(x-y)}+\frac{ y\cdot{y}(x\ +\ y)}{ xy(x+y)(x-y)}\right)\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\frac{ x^3-x^2y-(x^2y+y^3)+xy^2+y^3}{ xy(x+y)(x-y)}\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\frac{x^3-x^2y-x^2y-y^3+xy^2+y^3}{ xy(x+y)(x-y)}\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\frac{x^3-2x^2y+xy^2}{ xy(x+y)(x-y)}\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\frac{ x(x^2-2xy+y^2)}{ xy(x+y)(x-y)}\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\frac{ x(x-y)^2}{ xy(x+y)(x-y)}\cdot\frac{ (x+y)(x^2-xy+y^2)}{ (x-y)^2}=\frac{x^2-xy+y^2}{ y(x-y)}\)
Что и требовалось доказать.
- \({\largeв)}\ \left(\frac{2x^2y+2xy^2}{7x^3+x^2y+7xy^2+y^3}\cdot\frac{7x+y}{x^2-y^2}+\frac{x-y}{x^2+y^2}\right)\cdot(x^2-y^2)=\left(\frac{ 2xy(x+y)}{ 7x(x^2+y^2)+y(x^2+y^2)}\cdot\frac{7x+y}{ (x+y)(x-y)}+\frac{x-y}{x^2+y^2}\right)\cdot(x^2-y^2)=\left(\frac{ 2xy(x+y)}{ (x^2+y^2)(7x+y)}\cdot\frac{7x+y}{ (x+y)(x-y)}+\frac{x-y}{x^2+y^2}\right)\cdot(x^2-y^2)=\left(\frac{2xy^{\backslash1}}{ (x^2+y^2)(x-y)\phantom{^{\backslash1}}}+\frac{x-y^{\backslash{x\ -\ y}}}{x^2+y^2\phantom{^{\backslash{x\ -\ y}}}}\right)\cdot(x+y)(x-y)=\left(\frac{2xy}{ (x^2+y^2)(x-y)}+\frac{ (x-y)(x-y)}{ (x^2+y^2)(x-y)}\right)\cdot(x+y)(x-y)=\frac{ (2xy+x^2-2xy+y^2)(x+y)(x-y)}{ (x^2+y^2)(x-y)}=\frac{ (x^2+y^2)(x+y)(x-y)}{ (x^2+y^2)(x-y)}=x+y\)
Что и требовалось доказать.
- \({\largeг)}\ \left(\frac{5}{a^2-2a-ax+2x}-\frac{1}{8-8a+2a^2}\cdot\frac{20-10a}{x-2}\right):\frac{25}{x^3-8}=\left(\frac{5}{ a(a-x)-2(a-x)}-\frac{1}{ 2(4-4a+a^2)}\cdot\frac{ 10(2-a)}{x-2}\right)\cdot\frac{x^3-2^3}{25}=\left(\frac{5}{ (a-x)(a-2)}-\frac{1}{ 2(2-a)^2}\cdot\frac{ 10(2-a)}{x-2}\right)\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\left(\frac{5}{ (a-x)(a-2)}-\frac{5}{ (2-a)(x-2)}\right)\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\left(\frac{5^{\backslash{x\ -\ 2}}}{ (a-x)(a-2)}+\frac{5^{\backslash{a\ -\ x}}}{ (a-2)(x-2)}\right)\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\left(\frac{ 5(x-2)}{ (a-x)(a-2)(x-2)}+\frac{ 5(a-x)}{ (a-x)(a-2)(x-2)}\right)\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\frac{5x-10+5a-5x}{ (a-x)(a-2)(x-2)}\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\frac{5a-10}{ (a-x)(a-2)(x-2)}\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\frac{ 5(a-2)}{ (a-x)(a-2)(x-2)}\cdot\frac{ (x-2)(x^2+2x+4)}{25}=\frac{x^2+2x+4}{ 5(a-x)}\)
Что и требовалось доказать.
- \({\largeд)}\ \left(\frac{3a}{9-3x-3a+ax}-\frac{1}{a^2-9}:\frac{x-a}{3a^2+9a}\right)\cdot\frac{x^3-27}{3a}=\left(\frac{3a}{ 3(3-x)-a(3-x)}-\frac{1}{ (a+3)(a-3)}\cdot\frac{ 3a(a+3)}{x-a}\right)\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\left(\frac{3a}{ (3-x)(3-a)}-\frac{3a}{ (a-3)(x-a)}\right)\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\left(\frac{3a^{\backslash{x\ -\ a}}}{ (3-x)(3-a)}+\frac{3a^{\backslash{3\ -\ x}}}{ (3-a)(x-a)}\right)\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\left(\frac{ 3a(x-a)}{ (3-x)(3-a)(x-a)}+\frac{ 3a(3-x)}{ (3-x)(3-a)(x-a)}\right)\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\frac{3ax-3a^2+9a-3ax}{ (3-x)(3-a)(x-a)}\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\frac{9a-3a^2}{ (3-x)(3-a)(x-a)}\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\frac{ 3a(3-a)}{ {-}(x-3)(3-a)(x-a)}\cdot\frac{ (x-3)(x^2+3x+9)}{3a}=\frac{x^2+3x+9}{a-x}\)
Что и требовалось доказать.
