Question

Cho biểu thức A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right):\frac{4}{2\sqrt{x}+2}\)

Tìm x để A > 0

Answer 1

ĐKXĐ: \(x\ge0;x\ne1\)

\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}+1}{2}\right)\)

\(=\left(\frac{x-\sqrt{x}+x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}+1}{2}\right)=\frac{2x\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)

\(A>0\Leftrightarrow\frac{x}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)