Question

Упростить $\frac{\sqrt{x}+1}{1+\sqrt{x}+x}:\frac{1}{x^2-\sqrt{x}}$

Answer 1

$\frac{\sqrt{x}+1}{1+\sqrt{x}+x}:\frac{1}{x^2-\sqrt{x}}=\frac{\sqrt{x}+1}{1+\sqrt{x}+x} \cdot\frac{x^2-\sqrt{x}}{1}=\frac{\sqrt{x}+1}{1+\sqrt{x}+x}\cdot\frac{\sqrt{x}((\sqrt{x})^3-1)}{1}=\frac{\sqrt{x}+1}{1+\sqrt{x}+x}\cdot\frac{\sqrt{x}(\sqrt{x}-1)(1+\sqrt{x}+x)}{1} =\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)=\sqrt{x}(x-1)=x\sqrt{x}-\sqrt{x}.$