Chứng minh rằng: (a căn bậc hai của b +b) (căn bậc hai của a

* Hướng dẫn giải

Nhận xét rằng:

$\frac{\left(a\sqrt{b}+b\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\frac{\sqrt{b}\left(a\sqrt{b}+b\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}$

$$\begin{gathered} \sqrt{\frac{ab+b^2-2\sqrt{a{b}^3}}{a\left(a+2\sqrt{b}\right)+b}}=\sqrt{\frac{b\left[{\left(\sqrt{a}\right)}^2-2\sqrt{a}\sqrt{b}+{\left(\sqrt{b}\right)}^2\right]}{a^2+2a\sqrt{b}+{\left(\sqrt{b}\right)}^2}} \\ =\sqrt{\frac{b{\left(\sqrt{a}-\sqrt{b}\right)}^2}{{\left(a+\sqrt{b}\right)}^2}}=\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}} \end{gathered}$$

Do đó

$$\begin{gathered} VT=\frac{\sqrt{b}\left(a+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b}.\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}} \\ =\frac{b\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b} \\ =b \end{gathered}$$