$ĐKXĐ : $ $ a \neq 1, $ \(a\ge0\)
Ta có :
\(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}.\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1+\sqrt{a}-a\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left[\left(1-\sqrt{a}\right).\left(1+\sqrt{a}\right)\right]^2}\)
\(=\frac{\left(1+\sqrt{a}\right)-a.\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)}\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\frac{\left(1+\sqrt{a}\right).\left(1-a\right)}{\left(1-\sqrt{a}\right)}\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=1\)
ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
Ta có :
\(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)
\(=\left(a+2\sqrt{a}+1\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)
\(=\frac{\left(1+\sqrt{a}\right)^2\left(1-\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2\left(1-\sqrt{a}\right)^2}\)
\(=1\)
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