Question

помогиииите подробное решение пожалуйста

Answer 1

$1)\frac{x^2+4}{x\sqrt{\left (\frac{x^2-4}{2x}\right )^2+4}}=\frac{x^2+4}{x\sqrt{\frac{x^4-8x^2+16+4\cdot 4x^2}{4x^2}}}=\frac{x^2+4}{x\sqrt{\frac{x^4+8x^2+16}{4x^2}}}=\frac{x^2+4}{x\cdot {\frac{\sqrt{(x^2+4)^2}}{2|x|}}}=\\\\=\frac{(x^2+4)\cdot 2|x|}{x\cdot |x^2+4|}}=[\, |x^2+4|=x^2+4,\; t.k.\; x^2+4\ \textgreater \ 0\, ]=\\\\=\frac{2|x|}{x}= \left \{ {{2,\; esli\; x \geq 0,\; t.k.\; |x |=x} \atop {-2,\; esli\; x\ \textless \ 0,\; t.k.\; |x|=-x}} \right.$

$2)\; ...=\left (\frac{x+1}{x-1}\cdot \frac{\sqrt{x-1}}{\sqrt{x+1}}+\frac{x+1}{\sqrt{x-1}\cdot \sqrt{x+1}\right)}:\left (\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}-\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}\right )=\\\\=\left (\frac{\sqrt{x+1}}{\sqrt{x-1}}+\frac{\sqrt{x+1}}{\sqrt{x-1}\right):\frac{(x+\sqrt{x^2-1})^2-(x-\sqrt{x^2-1})^2}{x^2-(\sqrt{x^2-1})^2}}=\\\\=\frac{\sqrt{(x+1)(x-1)}+\sqrt{(x+1)(x-1)}}{(\sqrt{x-1})^2}:\\\\:\frac{x^2+2x\sqrt{x^2-1}+(x^2-1)-x^2+2x\sqrt{x^2-1}-(x^2-1)}{x^2-(x^2-1)}=$

$=\frac{2\sqrt{x^2-1}}{x-1}:\frac{4x\sqrt{x^2-1}}{1}=\frac{2\sqrt{x^2-1}}{(x-1)\cdot 4x\sqrt{x^2-1}}=\frac{1}{2x(x-1)}$