Question

Cho P = (\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)) : \(\dfrac{\sqrt{x}-1}{2}\)

a) Rút gọn biểu thức P

Answer 1

Lời giải:

ĐK: \(x\geq 0; x\neq 1\)

Ta có:

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

\(=\frac{x+2}{(\sqrt{x})^3-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

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

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

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

Do đó:

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