Question

\(\dfrac{4}{\sqrt{b}-1}+\dfrac{3}{\sqrt{b}+1}-\dfrac{6\sqrt{b}+2}{b-1}\)

Rút gọn

Answer 1

Ta có: \(\dfrac{4}{\sqrt{b}-1}+\dfrac{3}{\sqrt{b}+1}-\dfrac{6\sqrt{b}+2}{b-1}\) (ĐKXĐ: b ≥ 0, b ≠ 1)

\(=\dfrac{4\left(\sqrt{b}+1\right)+3\left(\sqrt{b}-1\right)-\left(6\sqrt{b}+2\right)}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}\)

\(=\dfrac{\sqrt{b}-1}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}\)

\(=\dfrac{1}{\sqrt{b}+1}\)