a,A\(=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\left(\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\left(\frac{1+\sqrt{x}}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{x-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b,Để A=\(\frac{1}{3}\)thì:
\(\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{1}{3}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{1}{3}=0\)
\(\Leftrightarrow\frac{3\sqrt{x}-3}{3\sqrt{x}}-\frac{\sqrt{x}}{3\sqrt{x}}=0\)
\(\Leftrightarrow3\sqrt{x}-3-\sqrt{x}=0\)
\(\Leftrightarrow2\sqrt{x}-3=0\)
\(\Leftrightarrow2\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{9}{4}\)
Lần sau nhớ viết đúng đầu bài nhé bạn!Mk giải luôn từ đầu cho đỡ mất thời gian!
\(\left(\sqrt{x-1}\right)^{^2}\)hay \(\left(\sqrt{x}-1\right)^2\)hả bạn?
