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 $\left ( \sqrt{6} +\sqrt{5}\right )^{2}-\sqrt{120}=11$