a) Biến đổi vế trái để được vế phải.
Ta có:
\(VT=\left ( \dfrac{1-a\sqrt{a}}{1-\sqrt{a}} +\sqrt{a}\right ). \left ( \dfrac{1-\sqrt{a}}{1-a} \right )^{2}\)
\(=\left ( \dfrac{1-(\sqrt{a})^3}{1-\sqrt{a}} +\sqrt{a}\right ). \left ( \dfrac{1-\sqrt{a}}{(1-\sqrt a)(1+ \sqrt a)} \right )^{2}\)
\(=\left ( \dfrac{(1-\sqrt{a})(1+\sqrt a+(\sqrt a)^2)}{1-\sqrt{a}} +\sqrt{a}\right ). \left ( \dfrac{1}{1+ \sqrt a} \right )^{2}\)
\(=\left [ (1+\sqrt a+(\sqrt a)^2) +\sqrt{a}\right ]. \dfrac{1}{(1+ \sqrt a)^2}\)
\(=\left [ (1+2\sqrt a+(\sqrt a)^2)\right ]. \dfrac{1}{(1+ \sqrt a)^2}\)
\(=(1+\sqrt a)^2. \dfrac{1}{(1+ \sqrt a)^2}=1=VP\).
b) Ta có:
\(VT=\dfrac{a+b}{b^{2}}\sqrt{\dfrac{a^{2}b^{4}}{a^{2}+2ab+b^{2}}}\)
\(=\dfrac{a+b}{b^{2}}\sqrt{\dfrac{(ab^2)^2}{(a+b)^2}}\)
\(=\dfrac{a+b}{b^{2}}\dfrac{\sqrt{(ab^2)^2}}{\sqrt{(a+b)^2}}\)
\(=\dfrac{a+b}{b^{2}}\dfrac{|ab^2|}{|a+b|}\)
\(=\dfrac{a+b}{b^{2}}.\dfrac{|a|b^2}{a+b}=|a|=VP\)
Vì \(a+b > 0 \Rightarrow |a+b|=a+b\).
