Question

Rút gọn biểu thức

A. \(\dfrac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(\dfrac{\sqrt{ }}{\sqrt{ }}\)

Answer 1

A = \(\dfrac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\) - \(\dfrac{a-b}{\sqrt{a+\sqrt{b}}}\)

= \(\dfrac{^{\left(\sqrt{a}+\sqrt{b}\right)^2}}{\sqrt{a}+\sqrt{b}}\) - \(\dfrac{a-b}{\sqrt{a+\sqrt{b}}}\)

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

= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a-b\right)}{\sqrt{a}+\sqrt{b}}\)

=\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)

=\(2\sqrt{b}\)