\(\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\dfrac{1+\sqrt{a}}{a-1}\right)^2\) (ĐK:\(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\))
\(=\left(\dfrac{1+a\sqrt{a}-\sqrt{a}-a}{1+\sqrt{a}}\right)\left(\dfrac{1+\sqrt{a}}{a-1}\right)^2\)
\(=\left(\dfrac{1-a+\sqrt{a}\left(a-1\right)}{1+\sqrt{a}}\right)\left(\dfrac{1+\sqrt{a}}{a-1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a-1\right)}{1+\sqrt{a}}\right)\left(\dfrac{1+\sqrt{a}}{a-1}\right)^2\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{a-1}=\dfrac{a-1}{a-1}=1\)(đpcm)
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