Question

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

a) Tìm ĐKXĐ :

b) Tìm giá trị của x để \(A=\frac{1}{3}\)

Answer 1

a) \(ĐKXĐ:0< x;x\ne1\)

b) Có : \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

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

Để \(A=\frac{1}{3}\) thì \(\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{1}{3}\Rightarrow\sqrt{x}=3\left(\sqrt{x}-1\right)\Leftrightarrow-2\sqrt{x}=-3\Leftrightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\left(TMĐKXĐ\right)\)

Vậy \(x=\frac{9}{4}\) thì \(A=\frac{1}{3}\)