Question

[Nguyễn Bảo Nhi]

rút gọn biểu thức

\(B=\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2\sqrt{x}}{1-x}\left(x\ge0,x\ne1\right)\)

Answer 1

[Nguyễn Huy Tú]

Với \(x\ge0;x\ne1\)

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

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