Question

Довести що √x+1/x√x+x+√x:x-1/x^2-√x=1

Answer 1

$\displaystyle\bf\\\frac{\sqrt{x} +1}{x\sqrt{x} +x+\sqrt{x} } :\frac{x-1}{x^{2} -\sqrt{x} }=\\\\\\=\frac{\sqrt{x} +1}{(x\cdot \sqrt{x} +(\sqrt{x} )^{2} +\sqrt{x} } :\frac{(\sqrt{x} )^{2} -1^{2} }{[(\sqrt{x} )^{2} ]^{2} -\sqrt{x} }=\\\\\\=\frac{\sqrt{x} +1}{\sqrt{x} \cdot(x+\sqrt{x} +1)} :\frac{(\sqrt{x} -1)\cdot(\sqrt{x} +1)}{\sqrt{x} \cdot[(\sqrt{x} )^{3}-1] } =$

$\displaystyle\bf\\=\frac{\sqrt{x} +1}{\sqrt{x} \cdot(x+\sqrt{x} +1)}\cdot\frac{\sqrt{x} \cdot(\sqrt{x} -1)\cdot(x+\sqrt{x} +1)}{(\sqrt{x} -1)\cdot(\sqrt{x} +1)} =1$