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- ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\\2\sqrt{a}\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}a\ne0\\a\ge0\\a\ne1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
- Ta có phương trình : \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(a-2\sqrt{a}+1-a-2\sqrt{a}-1\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(-4\sqrt{a}\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{-4a}{a-1}\right)\)= \(\frac{-4a\left(a-1\right)}{2\sqrt{a}\left(a-1\right)}\) = \(\frac{-4a}{2\sqrt{a}}\)
= \(\frac{-4\sqrt{a}\sqrt{a}}{2\sqrt{a}}\) = \(-2\sqrt{a}\)
\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)
....... Tới đây được chưa bạn?
ĐKXĐ:...
\(V=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1-2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\right)=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(W=\left(\frac{\sqrt{a}-1}{a+\sqrt{a}+1}-\frac{a-3\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{1}{\sqrt{a}-1}\right).\left(\frac{1-\sqrt{a}}{a+1}\right)\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2-a+3\sqrt{a}-1-\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{1-\sqrt{a}}{a+1}\right)\)
\(=\left(\frac{-\left(a+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{-\left(\sqrt{a}-1\right)}{a+1}\right)=\frac{1}{a+\sqrt{a}+1}\)
\(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\) (\(a>0;a\ne1\))
=\(\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\):\(\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
=\(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
=\(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\)
\(=\frac{a-1}{\sqrt{a}}\)
vậy...