Sửa đề: \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Bài này chắc là quy đồng full quá nhỉ?
a)P\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P =1/4 \(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow\sqrt{x}-2=\frac{3}{4}\sqrt{x}\)
\(\Leftrightarrow\frac{1}{4}\sqrt{x}=2\Leftrightarrow\sqrt{x}=8\Rightarrow x=64\left(TM\right)\)
P/s: Ko chắc..
Sửa đề :
a) \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) Để P \(=\frac{1}{4}\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}-8=3\sqrt{x}\Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\left(TM\right)\)