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1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)
2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\)
\(=2\cdot\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
a
\(ĐKXĐ:a\ne0;a\ne1;a\ne\sqrt{2}\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{1}\)
\(Q=\frac{\sqrt{a}-2}{\sqrt{a}}\)
b
\(Q>0\Leftrightarrow\frac{\sqrt{a}-2}{\sqrt{a}}>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>\sqrt{2}\)
a) ĐKXĐ: \(x,y\ge0\)
\(M=\dfrac{x\sqrt{y}-\sqrt{y}-y\sqrt{x}+\sqrt{x}}{1+\sqrt{xy}}=\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1+\sqrt{xy}}=\sqrt{x}-\sqrt{y}\)
b) \(x=\left(1-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
\(y=3-\sqrt{8}\Rightarrow\sqrt{y}=\sqrt{3-\sqrt{8}}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
\(\Rightarrow M=\left(\sqrt{3}-1\right)-\left(\sqrt{2}-1\right)=\sqrt{3}-\sqrt{2}\)
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