Bài Làm:
1, \(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(=\frac{x+2}{\left(\sqrt{x}\right)^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
2, Ta có: \(x=4-2\sqrt{3}\)
\(\Rightarrow\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{3-2.\sqrt{3}.1+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow A=\frac{\sqrt{3}-1}{4-2\sqrt{3}+\sqrt{3}-1+1}=\frac{\sqrt{3}-1}{4-\sqrt{3}}=\frac{\left(\sqrt{3}-1\right)\left(4+\sqrt{3}\right)}{4^2-\left(\sqrt{3}\right)^2}=\frac{3\sqrt{3}-1}{13}\)Chúc pạn hok tốt!!!
1. A = \(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
= \(\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
= \(\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
2. x = \(4-2\sqrt{3}\)
= 3 - \(2\sqrt{3}\) + 1
= \(\left(\sqrt{3}-1\right)^2\)
=> \(\sqrt{x}=\sqrt{3}-1\)
=> A = \(\frac{\sqrt{3}-1}{4-2\sqrt{3}+\sqrt{3}-1+1}\)
= \(\frac{\sqrt{3}-1}{3-\sqrt{3}+1}\)
