Упражнение 58. Даю 35 балла. Created by VALERA123321456. algebra-ru

Упражнение 58. Даю 35 балла...

VALERA123321456 @VALERA123321456

October 2021 1 16 Report

Упражнение 58. Даю 35 балла

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NNNLLL54

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$1)\; \; \frac{x}{a+c}=\frac{c}{c^2-a^2}\; ,\; \; x=\frac{c\, (a+c)}{(a-c)(a+c)}=\frac{c}{a-c}\\\\2)\; \; \frac{x}{2n+c}=\frac{4c}{c^2-4n^2}\; ,\; \; x=\frac{4c\, (2n+c)}{(c-2n)(c+2n)}=\frac{4c}{c-2n}\\\\3)\; \; \frac{x}{a-3c}=\frac{2c}{9c^2-a^2}\; ,\; \; x=\frac{2c\, (a-3c)}{(3c-a)(3c+a)}=-\frac{2c}{3c+a}\\\\4)\; \; \frac{6c}{25c^2-4a^2}=\frac{x}{5c+2a}\; ,\; \; x=\frac{6c\, (5c+2a)}{(5c-2a)(5c+2a)}=\frac{6c}{5c-2a}$

$5)\; \; \frac{a-x}{2a-3c}=\frac{3ac}{9c^2-4a^2}\; ,\; \; a-x=\frac{3ac\, (2a-3c)}{(3c-2a)(3c+2a)}=-\frac{3ac}{3c+2a}\; ,\\\\x=a+\frac{3ac}{3c+2a}=\frac{6ac+2a^2}{3c+2a}\\\\6)\; \; \frac{2a-x}{3b-a}=\frac{6ab}{9b^2-a^2}\; ,\; \; 2a-x=\frac{6ab\, (3b-a)}{(3b-a)(3b+a)}=\frac{6ab}{3b-a}\; ,\\\\x=2a-\frac{6ab}{3b-a}=\frac{-2a^2}{3b-a}$

P.S. Из основного свойства пропорции следуют равенства:

$\frac{a}{b}=\frac{c}{d}\; \; \Rightarrow \; \; a=\frac{b\cdot c}{d}\; \; ,\; \; c=\frac{a\cdot d}{c}\; \; ,\; \; b=\frac{a\cdot d}{c}\; \; ,\; \; d=\frac{b\cdot c}{a}$

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