Question

rút gọn biểu thức A
A = \(\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\) với x . 0 vào x \(\ne\) 1

Answer 1

\(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}ĐK:x\ne1;x\ne0\)

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

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

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

\(=\sqrt{x}-1\)

Answer 2

Lời giải:

Ta có:

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

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