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October 2021 1 11 Report

Алгебра, 8 класс. 99 баллов.
1. [tex] \frac{a-3}{a+3} - \frac{a+2}{a-2} [/tex]
2.[tex] \frac{p+2}{p+1} - \frac{p+6}{p-3} [/tex]
3.[tex] \frac{x+4y}{5y(x+y)}- \frac{x-y}{5y(x-4y)} [/tex]
4.[tex] \frac{d-c}{3c(2c+d)} + \frac{2c-d}{3c(c+d)} [/tex]

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90misha90

Verified answer

1
$\frac{a-3}{a+3}+ \frac{a+2}{a-2}= \frac{(a-3)(a-2)-(a+3)(a+2)}{(a+3)(a-2)} = \frac{a^2-2a-3a+6-(a^2+2a+3a+6)}{a^2-2a+3a-6} =$
$=\frac{-10a}{a^2+a-6} = -\frac{10a}{a^2+a-6} .$
2
$\frac{p+2}{p+1} - \frac{p+6}{p-3} = \frac{(p+2)(p-3)-(p+6)(p+1)}{(p+1)(p-3)} = \frac{p^2-p-6-(p^2+7p+6)}{p^2-2p-3} = -\frac{8p}{p^2-2p-3}$
3
$\frac{1}{5y}*( \frac{x+4y}{x+y}- \frac{x-y}{x-4y} )= \frac{1}{5y}* \frac{(x+4y)(x-4y)-(x-y)(x+y)}{(x+y)(x-4y)} =$
$= \frac{1}{5y} * \frac{x^2-(4y)^2-(x^2-y^2)}{x^2-4xy+yx-4y^2} = \frac{1}{5y} * \frac{-15y^2}{x^2-3xy-4y^2} =- \frac{3y}{x^2-3xy-4y^2}$
4
$\frac{1}{3c} *( \frac{d-c}{2c+d} + \frac{2c-d}{c+d} )= \frac{1}{3c} * \frac{(d-c)(d+c)+(2c-d)(2c+d)}{(2c+d)(c+d)} = \frac{1}{3c}* \frac{d^2-c^2+4c^2-d^2}{2c^2+3cd+d^2} =$
$=\frac{1}{3c}* \frac{3c^2}{2c^2+3cd+d^2}= \frac{c}{2c^2+3cd+d^2}$

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