Question

Bài 4. Trục căn thức ở mẫu của các biểu thức sau:

a) \(\dfrac{10 - 3\sqrt{5}}{\sqrt{5}}\).

b) \(\dfrac{2\sqrt{3}}{\sqrt{3} - 2}\).

c) \(\dfrac{12\sqrt{2}}{\sqrt{8} + 2}\).

d) \(\dfrac{\sqrt{10} + \sqrt{15}}{\sqrt{10} - \sqrt{15}}\).

Answer 1

\(a.\frac{10-3\sqrt5}{\sqrt5}=\frac{10}{\sqrt5}-\frac{3\sqrt5}{\sqrt5}=\frac{10}{\sqrt5}-3\)

\(=\frac{10\sqrt5}{5}-3=2\sqrt5-3\)

\(b.\frac{2\sqrt3}{\sqrt3-2}=\frac{2\sqrt3}{\sqrt3-2}\cdot\frac{\sqrt3+2}{\sqrt3+2}=\frac{2\sqrt3\cdot\left(\sqrt3+2\right)}{\left(\sqrt3\right)^2-2^2}\)

\(=\frac{2\sqrt3\cdot\left(\sqrt3+2\right)}{-1}=-2\sqrt3\cdot\left(\sqrt3+2\right)=-6-4\sqrt3\)

\(c.\frac{12\sqrt2}{\sqrt8+2}=\frac{12\sqrt2}{2\sqrt2+2}=\frac{12\sqrt2}{2\cdot\left(\sqrt2+1\right)}=\frac{6\sqrt2}{\sqrt2+1}\)

\(=\frac{6\sqrt2}{\sqrt2+1}\cdot\frac{\sqrt2-1}{\sqrt2-1}=\frac{6\sqrt2\cdot\left(\sqrt2-1\right)}{2-1}=6\sqrt2\cdot\left(\sqrt2-1\right)=12-6\sqrt2\)

\(d.\frac{\sqrt{10}+\sqrt{15}}{\sqrt{10}-\sqrt{15}}=\frac{\sqrt{10}+\sqrt{15}}{\sqrt{10}-\sqrt{15}}\cdot\frac{\sqrt{10}+\sqrt{15}}{\sqrt{10}+\sqrt{15}}=\frac{\left(\sqrt{10}+\sqrt{15}\right)^2}{10-15}\)

\(=\frac{10+15+2\sqrt{150}}{-5}=\frac{25+10\sqrt6}{-5}=-5-2\sqrt6\)