Question

Đề bài: Cho biểu thức :

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

a) Rút gọn M.

b) Tính giá trị của M khi \(x=7+4\sqrt{3}\)

c) Tìm \(x\) sao cho M \(=\frac{1}{2}\)

Giúp em với ạ !

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

b) Khi \(x=7+4\sqrt{3}\Rightarrow\frac{7+4\sqrt{3}}{\sqrt{\left(2+\sqrt{3}\right)^2}+1}=\frac{7+4\sqrt{3}}{3+\sqrt{3}}\)

c)\(M=\frac{1}{2}\Leftrightarrow\frac{x}{\sqrt{x}+1}=\frac{1}{2}\Leftrightarrow\sqrt{x}=2x-1\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x^2=4x^2-4x+1\Leftrightarrow3x^2-4x+1=0\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=\frac{1}{3}\left(l\right)\\x=1\left(l\right)\end{matrix}\right.\end{matrix}\right.\)