Question

Rút gọn biểu thức chứa căn thức bậc hai:
a,\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)với a\(\ge\)0,b\(\ge\)0 và a \(\ne\)b
b,\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\) điều kiện giống câu trên

Answer 1

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

=\(\frac{a+b+2\sqrt{ab}+a+b-2\sqrt{ab}}{a-b}=\frac{2\left(a+b\right)}{a-b}\)

b/\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

=\(\sqrt{a}+\sqrt{b}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)=\(\frac{a+b+2\sqrt{ab}+a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}=\frac{2a+2b+3\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)