Giải bài 33 trang 66

Đề bài

Trục căn thức ở mẫu:

a) $\frac{2-\sqrt{5}}{\sqrt{5}}$

b) $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

c) $\frac{8}{3\sqrt{5}+3}$

d) $\frac{1}{\sqrt[3]{33}+\sqrt[3]{37}}$

Lời giải chi tiết

a) $\frac{2-\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}\left(2-\sqrt{5}\right)}{\sqrt{5}.\sqrt{5}}=\frac{\sqrt{5}\left(2-\sqrt{5}\right)}{5}$

b) $\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{{\left(\sqrt{2}+1\right)}^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}={\left(\sqrt{2}+1\right)}^2$

c) $\frac{8}{3\sqrt{5}+3}=\frac{8}{3\left(\sqrt{5}+1\right)}$

$=\frac{8\left(\sqrt{5}-1\right)}{3\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8\left(\sqrt{5}-1\right)}{3\left(5-1\right)}=\frac{2\left(\sqrt{5}-1\right)}{3}$

d) $\frac{1}{\sqrt[3]{33}+\sqrt[3]{37}}=\frac{{\left(\sqrt[3]{33}\right)}^2-\sqrt[3]{33}.\sqrt[3]{37}+{\left(\sqrt[3]{37}\right)}^2}{\left(\sqrt[3]{33}+\sqrt[3]{37}\right)\left[{\left(\sqrt[3]{33}\right)}^2-\sqrt[3]{33}.\sqrt[3]{37}+{\left(\sqrt[3]{37}\right)}^2\right]}$

$=\frac{\sqrt[3]{33^2}-\sqrt[3]{33.7}+\sqrt[3]{37^2}}{{\left(\sqrt[3]{33}\right)}^3+{\left(\sqrt[3]{37}\right)}^3}=\frac{\sqrt[3]{39}-\sqrt[3]{321}+\sqrt[3]{349}}{10}$