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#)Giải :
Bài 1 :
a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\frac{\left(1-x\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x+1}\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) Để \(P>0\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow0< x< 1}\)
c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu ''='' xảy ra khi \(x=\frac{1}{4}\)
a) ĐK: \(x>0;x\ne1\)
\(P=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(=\left(\frac{\left(\sqrt{x}-1\right)^2}{x-1}-\frac{\left(\sqrt{x}+1\right)^2}{x-1}\right)\left(\frac{1-x}{2\sqrt{x}}\right)^2\)
\(=\frac{-4\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{4x}\)
\(=\frac{1-x}{\sqrt{x}}\)
\(M=\left(\frac{x+2}{x\sqrt{x-1}}+\frac{\sqrt{x}}{x+\sqrt{x+1}}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}-1}{7}\)
\(=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}-1}{7}\)
\(=\left[\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{7}\)
\(=\left[\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{7}\)
\(=\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{7}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2.7}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(=\frac{7}{x+\sqrt{x}+1}\)
Sửa lại đề nha , đề đúng nè :
\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x-1}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x+2\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{\left(\sqrt{x}+1\right)^2}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{x-\sqrt{x}-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}-x}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
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