Giải câu 62 bài: Luyện tập sgk Toán 9 tập 1 Trang 33.
Ta có :
a. $\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}$
= $\frac{1}{2}\sqrt{4^{2}.3}-2\sqrt{5^{2}.3}-\sqrt{\frac{33}{11}}+5\sqrt{\frac{4.3}{3^{2}}}$
= $2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10\sqrt{3}}{3}$
= $-9\sqrt{3}+\frac{10\sqrt{3}}{3}=\frac{-27\sqrt{3}}{3}+\frac{10\sqrt{3}}{3}=\frac{-17\sqrt{3}}{3}$
Vậy $\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}=\frac{-17\sqrt{3}}{3}$
b. $\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\frac{2}{3}}-\sqrt{6}$
= $\sqrt{5^{2}.6}+\sqrt{4^{2}.6}+4,5.\sqrt{\frac{8.3}{3^{2}}}-\sqrt{6}$
= $5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}$
Vậy $\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\frac{2}{3}}-\sqrt{6}=11\sqrt{6}$
c. $\left ( \sqrt{28}-2\sqrt{3}+\sqrt{7} \right )\sqrt{7}+\sqrt{84}$
= $\left ( \sqrt{2^{2}.7}-2\sqrt{3}+\sqrt{7} \right )\sqrt{7}+\sqrt{2^{2}.21}$
= $\left ( 2\sqrt{7}-2\sqrt{3}+\sqrt{7} \right )\sqrt{7}+2\sqrt{21}$
= $14-2\sqrt{21}+7+2\sqrt{21}=21$
Vậy $\left ( \sqrt{28}-2\sqrt{3}+\sqrt{7} \right )\sqrt{7}+\sqrt{84}=21$
d. $\left ( \sqrt{6} +\sqrt{5}\right )^{2}-\sqrt{120}$
= $6+2\sqrt{30}+5-\sqrt{2^{2}.30}$
= $11+2\sqrt{20}-2\sqrt{20}=11$
Vậy $\left ( \sqrt{6} +\sqrt{5}\right )^{2}-\sqrt{120}=11$