№ 16.6 Алгебра = № 32.6 Математика
Обчисліть:
1. $(-\sqrt{11})^2;$
2. $\sqrt{19}\cdot\sqrt{19};$
3. $(2\sqrt7)^2;$
4. $\left(-\frac{1}{4}\sqrt8\right)^2;$
5. $-7\sqrt3\cdot\sqrt3;$
6. $0{,}2\cdot(-\sqrt5)^2;$
7. $\left(\frac{1}{\sqrt{15}}\right)^2;$
8. $\left(-\frac{\sqrt{10}}{3}\right)^2.$
Розв'язок:
1. $(-\sqrt{11})^2=(\sqrt{11})^2=11;$
2. $\sqrt{19}\cdot\sqrt{19}=(\sqrt{19})^2=19;$
3. $(2\sqrt7)^2=4\cdot(\sqrt7)^2=$
$= 4\cdot7=28;$
4. $\left(-\frac{1}{4}\sqrt8\right)^2=\frac{1}{16}(\sqrt8)^2=$
$= \frac{1}{16}\cdot8=\frac{8}{16}=\frac{1}{2};$
5. $-7\sqrt3\cdot\sqrt3=-7\cdot(\sqrt3)^2=$
$= -7\cdot3=-21;$
6. $0{,}2\cdot(-\sqrt5)^2=0{,}2\cdot(\sqrt5)^2=$
$= 0{,}2\cdot5=1;$
7. $\left(\frac{1}{\sqrt{15}}\right)^2=\frac{1}{(\sqrt{15})^2}=\frac{1}{15};$
8. $\left(-\frac{\sqrt{10}}{3}\right)^2=\left(\frac{\sqrt{10}}{3}\right)^2=$
$= \frac{(\sqrt{10})^2}{9}=\frac{10}{9}=1\frac{1}{9}.$