Question

Помоги пожалуйста срочно с Алгеброй!!!

Answer 1

Ответ:

См. объяснение

Объяснение:

$\frac{6}{x+1} - \frac{10}{(1-x)(1+x)} +1 = \frac{5}{x-1} \\\frac{6}{x+1} + \frac{10}{(x-1)(x+1)} +1 = \frac{5}{x-1} \\\frac{6(x-1)}{(x-1)(x+1)} + \frac{10}{(x-1)(x+1)} +\frac{(x-1)(x+1)}{(x-1)(x+1)} = \frac{5(x+1)}{(x-1)(x+1)}\\\\6(x-1)+10+(x-1)(x+1)=5(x+1) и x\neq 1 и x\neq -1\\6x-6+10+x^{2}-1=5x+5 и x\neq 1 и x\neq -1\\x^{2}+x-2=0 и x\neq 1 и x\neq -1\\(x-1)(x+2)=0 и x\neq 1 и x\neq -1\\\\x=-2$

Область: D=R\{-1;1}

Answer 2

Ответ:

$\dfrac{6}{x+1}-\dfrac{10}{1-x^2}+1=\dfrac{5}{x-1}\\\\\\OOF:\left\{\begin{array}{l}x+1\ne 0\\1-x^2\ne 0\\x-1\ne 0\end{array}\right\ \ \left\{\begin{array}{l}x\ne -1\\(1-x)(1+x)\ne 0\\x\ne 1\end{array}\right\ \ \left\{\begin{array}{l}x\ne -1\\x\ne 1\ ,\ x\ne -1\\x\ne 1\end{array}\right\ \ \ \Rightarrow \\\\\\x\ne \pm 1\ \ \ \Rightarrow \ \ \ x\in R/\{-1;1\}\\\\\dfrac{6}{x+1}+\dfrac{10}{x^2-1}+1=\dfrac{5}{x-1}\ |\cdot (x-1)(x+1)\\\\\\6\, (x-1)+10+(x^2-1)=5(x+1)\\\\6x-6+10+x^2-1=5x+5\\\\x^2+x-2=0\ \ ,$

$x_1=-2\ ,\ \ x_2=1\ \ (po\ teoreme\ Vieta)\\\\x_2=1\notin OOF\ \ \ \Rightarrow \\\\Otvet:\ \ x=-2\ .$