Question

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

Tim cac gia tri cua x de A co nghia

Rut gon bieu thuc A

Answer 1

A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

Ta có:

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

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

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

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

Kết luận: ...

Answer 2

ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...

Answer 3