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$DK: x\neq 0$$\Leftrightarrow (x+1)\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}+(x-1)\sqrt{\frac{2}{x^2}-\frac{2}{x}+1}-5=0$$\Leftrightarrow (x+1)[\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}-(\frac{2}{x}+\frac{1}{2})]+(x-1)[\sqrt{\frac{2}{x^2}-\frac{2}{x}+1}-(\frac{2}{x}-\frac{1}{2})=0$$\Leftrightarrow (\frac{3}{4}-\frac{2}{x^2})$ $(.......)$=0$\Rightarrow \frac{2}{x^2}=\frac{3}{4}\Leftrightarrow x=\pm \sqrt{\frac{8}{3}}$hoặc đỏ=0 ( loại)

$DK: x\neq 0$$\Leftrightarrow (x+1)\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}+(x-1)\sqrt{\frac{2}{x^2}-\frac{2}{x}+1}-5=0$$\Leftrightarrow (x+1)[\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}-(\frac{2}{x}+\frac{1}{2})]+(x-1)[\sqrt{\frac{2}{x^2}-\frac{2}{x}+1}-(\frac{2}{x}-\frac{1}{2})=0$$\Leftrightarrow (\frac{3}{4}-\frac{2}{x^2})$ $(.......)$=0$\Rightarrow \frac{2}{x^2}=\frac{3}{4}\Leftrightarrow x=\pm \sqrt{\frac{8}{3}}=\pm \frac{2\sqrt{6}}{3}$hoặc đỏ=0 ( loại)