Question

trục căn thức ở mẫu

a.\(\dfrac{5}{3\sqrt{8}}\) , \(\dfrac{2}{\sqrt{b}}\) với b>0

b.\(\dfrac{5}{5-2\sqrt{3}}\), \(\dfrac{2a}{1-\sqrt{a}}\) với a\(\ge\)0 và a\(\ne\)1

c. \(\dfrac{4}{\sqrt{7}+\sqrt{5}}\) , \(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}\) với a>b>0

Answer 1

a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)

\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)

b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)

\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)

c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)

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