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\(A=\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}}{2x-1}-1\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{2x}-1+2x-2x+1}{2x-1}=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}\)
\(B=\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=1+\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{2x-1}\)
\(=1+\dfrac{-2\sqrt{x}-1-2x}{2x-1}\)
\(=\dfrac{2x-1-2\sqrt{x}-1-2x}{2x-1}=\dfrac{-2-2\sqrt{x}}{2x-1}\)
\(P=A:B=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}:\dfrac{-2\sqrt{x}-2}{2x-1}\)
\(=\dfrac{2\sqrt{x}\left(x+\sqrt{2}\right)}{2x-1}\cdot\dfrac{2x-1}{-2\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}\left(x+\sqrt{2}\right)}{\sqrt{x}+1}\)
b: Thay \(\sqrt{x}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\) vào P, ta được:
\(P=\left[-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\left(\dfrac{3+2\sqrt{2}}{2}+\sqrt{2}\right)\right]:\left[\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}+1\right]\)
\(=\left[\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\dfrac{3+4\sqrt{2}}{2}\right]:\left[\dfrac{2+\sqrt{2}+2}{2}\right]\)
\(=\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{4}\cdot\dfrac{2}{4+\sqrt{2}}\)
\(=\dfrac{-\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{2\cdot\left(2\sqrt{2}+1\right)}=\dfrac{-\left(4\sqrt{2}+3\right)}{3\cdot\left(3+\sqrt{2}\right)}\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\div\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}-\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\div\left[-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right]\)
\(=-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left[\left(2\sqrt{x}-1\right)\left(-\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\right)\right]\)
\(=-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left[\dfrac{-\left(x-\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\times\left(2\sqrt{x}-1\right)\right]\)
\(=-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\dfrac{-\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(=-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\dfrac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{-\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
Bài 1:
a: ĐKXĐ: x>0; x<>1
b: \(M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
c: \(M=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/4
a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
a) \(x>0,x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
c) \(P< 0\Rightarrow\dfrac{x-1}{\sqrt{x}}< 0\) mà \(\sqrt{x}>0\Rightarrow x-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)