Giải bài 1 trang 65 vở thực hành Toán 9

Đề bài

Rút gọn các biểu thức sau:

a) $\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{1}{\sqrt{5}-2}$;

b) $\sqrt{{\left(\sqrt{7}-2\right)}^2}-\sqrt{63}+\frac{\sqrt{56}}{\sqrt{2}}$;

c) $\frac{\sqrt{{\left(\sqrt{3}+\sqrt{2}\right)}^2}+\sqrt{{\left(\sqrt{3}-\sqrt{2}\right)}^2}}{2\sqrt{12}}$;

d) $\frac{\sqrt[3]{3{\left(\sqrt{2}+1\right)}^3}-1}{\sqrt{50}}$.

Lời giải chi tiết

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

$=\sqrt{5}+3-\frac{\sqrt{5}+2}{5-4}=1$

b) $\sqrt{{\left(\sqrt{7}-2\right)}^2}-\sqrt{63}+\frac{\sqrt{56}}{\sqrt{2}}$

$=\left|\sqrt{7}-2\right|-\sqrt{9.7}+\sqrt{\frac{56}{2}}=\sqrt{7}-2-3\sqrt{7}+2\sqrt{7}=-2;$

c) $\frac{\sqrt{{\left(\sqrt{3}+\sqrt{2}\right)}^2}+\sqrt{{\left(\sqrt{3}-\sqrt{2}\right)}^2}}{2\sqrt{12}}=\frac{\sqrt{3}+\sqrt{2}+\left|\sqrt{3}-\sqrt{2}\right|}{2.2\sqrt{3}}$

$=\frac{\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{4\sqrt{3}}=\frac{1}{2};$

d) $\frac{\sqrt[3]{3{\left(\sqrt{2}+1\right)}^3}-1}{\sqrt{50}}=\frac{\sqrt{2}+1-1}{5\sqrt{2}}=\frac{1}{5}$