rút gọn biểu thức A=1/căn x+1 +1/căn x-1
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a: ĐKXĐ: x>=0; x<>1
b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)
d: căn x+2>=2
=>A<=1/2
Dấu = xảy ra khi x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\cdot\dfrac{\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\left(\sqrt{x}+1\right)^2\)
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Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2\left(x+\sqrt{x}\right)}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x+\sqrt{x}-2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x+\sqrt{x}\)
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a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: Khi x=9/4 thì A=3/2:1/2=3/2*2=3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}-1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)
\(=\dfrac{2\sqrt{x}}{x-1}\)