Giải câu 58 bài 8: Rút gọn biểu thức chứa căn thức bậc hai sgk Toán 9 tập 1 Trang 32.
Ta có :
a. $5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20+\sqrt{5}}$
= $5\sqrt{\frac{5}{5^{2}}}+\frac{1}{2}\sqrt{2^{2}.5}+\sqrt{5}$
= $\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}$
Vậy $5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20+\sqrt{5}}=3\sqrt{5}$
b. $\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}$
= $\sqrt{\frac{1}{2}}+\sqrt{\frac{9}{2}}+\sqrt{\frac{25}{2}}$
= $\sqrt{\frac{1}{2}}+3\sqrt{\frac{1}{2}}+5\sqrt{\frac{1}{2}}=9\sqrt{\frac{1}{2}}$
Vậy $\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=9\sqrt{\frac{1}{2}}$
c. $\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}$
= $\sqrt{2^{2}.5}-\sqrt{3^{2}.5}+3\sqrt{3^{2}.2}+\sqrt{6^{2}.2}$
= $2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=15\sqrt{2}-\sqrt{5}$
Vậy $\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=15\sqrt{2}-\sqrt{5}$
d. $0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}$
= $0,1\sqrt{10^{2}.2}+2\sqrt{0,2^{2}.2}+0,4\sqrt{5^{2}.2}$
= $\sqrt{2}+0,4\sqrt{2}+2\sqrt{2}=3,4\sqrt{2}$
Vậy $0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}=3,4\sqrt{2}$