Question

$\frac{a-x}{b-a} -\frac{x+a}{a-b} = \frac{2ax}{a^2-b^2}$

Answer 1

Объяснение:

$\frac{a - x}{b - a} - \frac{x + a}{a - b} = \frac{2ax}{ {a}^{2} - {b}^{2} } \\ \frac{a - x + x + a}{b - a} = \frac{2ax}{ {a}^{2} - {b}^{2} } \\ 2a(a - b)(a + b) = 2ax(b - a) \\ 2a(a + b) = - 2ax \\ - x = a + b$

Answer 2

$\frac{a-x}{b-a}-\frac{x+a}{a-b}=\frac{2ax}{a^{2}-b^{2}} \\\\\frac{a-x}{b-a}+\frac{x+a}{b-a}=\frac{2ax}{a^{2}-b^{2}}\\\\\frac{a-x+x+a}{b-a}=\frac{2ax}{a^{2}-b^{2}}\\\\\frac{2a}{b-a}-\frac{2ax}{(a-b)(a+b)}=0\\\\\frac{2a}{b-a}+\frac{2ax}{(b-a)(a+b)}=0\\\\\frac{2a^{2}+2ab+2ax}{(b-a)(a+b)}=0\\\\\left \{ {{2a^{2}+2ab+2ax=0 } \atop {b-a\neq0;a+b\neq0}} \right. \\\\2ax=-2a^{2}-2ab\\\\x=-\frac{2a^{2}+2ab}{2a}=-(a+b)$