Giải SBT Toán 9 CTST Bài 3. Tính chất của phép khai phương

a) $\sqrt{5}\cdot \sqrt{11}=\sqrt{5\cdot 11}=\sqrt{55}.$

b) $\sqrt{\frac{10}{3}}\cdot \sqrt{\frac{3}{5}}=\sqrt{\frac{10}{3}\cdot \frac{3}{5}}=\sqrt{2}.$

c) $\sqrt{3}\cdot \sqrt{5}\cdot \sqrt{6}=\sqrt{3\cdot 5\cdot 6}=\sqrt{90}.$

d) $\sqrt{\frac{6}{7}}\cdot \sqrt{2{,}8}=\sqrt{\frac{6}{7}}\cdot \sqrt{\frac{14}{5}}=\sqrt{\frac{6}{7}\cdot \frac{14}{5}}=\sqrt{\frac{12}{5}}=\sqrt{2{,}4}$

a) $\sqrt{3\cdot {8}^{2}}=|8|\sqrt{3}=8\sqrt{3}.$

b) $\sqrt{150}=\sqrt{{5}^{2}\cdot 6}=|5|\sqrt{6}=5\sqrt{6}.$

c) $\sqrt{1\ 000}=\sqrt{{10}^{2}\cdot 10}=|10|\sqrt{10}=10\sqrt{10}.$

d) $\sqrt{{2}^{2}\cdot {5}^{4}\cdot 7}=\sqrt{{2}^{2}\cdot {({5}^{2})}^{2}\cdot 7}=\sqrt{(2\cdot {5}^{2})\cdot 7}=|2\cdot {5}^{2}|\sqrt{7}=2\cdot {5}^{2}\sqrt{7}=50\sqrt{7}.$

a) $6\sqrt{5}=\sqrt{{6}^{2}\cdot 5}=\sqrt{36\cdot 5}=\sqrt{180}.$

b) $-8\sqrt{10}=-\sqrt{{8}^{2}\cdot 10}=-\sqrt{64\cdot 10}=-\sqrt{640}.$

c) $5\sqrt{\frac{2}{5}}=\sqrt{{5}^{2}\cdot \frac{2}{5}}=\sqrt{10}.$

d) Với a ≥ 0; b > 0, ta có:

a) $\sqrt{\frac{16}{121}}=\sqrt{\frac{{4}^{2}}{{11}^{2}}}=\frac{\sqrt{{4}^{2}}}{\sqrt{{11}^{2}}}=\frac{4}{11}.$

b) $\sqrt{4\frac{21}{25}}=\sqrt{\frac{121}{25}}=\sqrt{\frac{{11}^{2}}{{5}^{2}}}=\frac{\sqrt{{11}^{2}}}{\sqrt{{5}^{2}}}=\frac{11}{5}.$

c) $\sqrt{\frac{6{,}4}{8{,}1}}=\sqrt{\frac{6{,}4\cdot 10}{8{,}1\cdot 10}}=\sqrt{\frac{64}{81}}=\sqrt{\frac{{8}^{2}}{{9}^{2}}}=\frac{\sqrt{{8}^{2}}}{\sqrt{{9}^{2}}}=\frac{8}{9}.$

d)$\sqrt{\frac{300}{27}}=\sqrt{\frac{100\cdot 3}{9\cdot 3}}=\sqrt{\frac{100}{9}}=\sqrt{\frac{{10}^{2}}{{3}^{2}}}=\frac{\sqrt{{10}^{2}}}{\sqrt{{3}^{2}}}=\frac{10}{3}.$

e) $\frac{\sqrt{6}}{\sqrt{150}}=\sqrt{\frac{6}{150}}=\sqrt{\frac{6}{25\cdot 6}}=\sqrt{\frac{1}{25}}=\sqrt{{(\frac{1}{5})}^{2}}=\frac{1}{5}$

g) $\sqrt{\frac{3}{2}}:\sqrt{\frac{1}{24}}=\sqrt{\frac{3}{2}:\frac{1}{24}}=\sqrt{\frac{3}{2}\cdot 24}=\sqrt{36}=\sqrt{{6}^{2}}=6.$