№ 16.16 Алгебра = № 32.16 Математика
Обчисліть:
1. $\left((-\sqrt7)^2\right)^2;$
2. $(3\sqrt7)^2-(7\sqrt3)^2;$
3. $16\cdot\left(-\frac{1}{2}\sqrt7\right)^2+\frac{1}{3}\cdot(4\sqrt3)^2;$
4. $\sqrt{70{,}56}-\left(\frac{1}{2}\sqrt{42}\right)^2;$
5. $(5\sqrt2)^2-5\cdot(-\sqrt2)^2;$
6. $\left(\frac{2}{3}\sqrt{\frac{9}{10}}\right)^2+\left(-\frac{5}{6}\sqrt{\frac{36}{65}}\right)^2.$
Розв'язок:
1. $\left((-\sqrt7)^2\right)^2=7^2=49;$
2. $(3\sqrt7)^2-(7\sqrt3)^2=$
$= 9·7-49·3=$
$= 7·3·(3-7)=$
$= 21·(-4)=-84;$
3. $16\cdot\left(-\frac{1}{2}\sqrt7\right)^2+\frac{1}{3}\cdot(4\sqrt3)^2=$
$= 16\cdot\frac{1}{4}\cdot7+\frac{1}{3}\cdot16\cdot3=$
$= 4·7+16=4·(7+4)=$
$= 4·11=44;$
4. $\sqrt{70{,}56}-\left(\frac{1}{2}\sqrt{42}\right)^2=$
$= 8{,}4-\frac{1}{4}\cdot42=8{,}4-\frac{21}{2}=$
$= 8{,}4-10{,}5=-2{,}1;$
5. $(5\sqrt2)^2-5\cdot(-\sqrt2)^2=$
$= 25\cdot2-5\cdot2=$
$= 50-10=40;$
6. $\left(\frac{2}{3}\sqrt{\frac{9}{10}}\right)^2+\left(-\frac{5}{6}\sqrt{\frac{36}{65}}\right)^2=$
$= \frac{4}{9}\cdot\frac{9}{10}+\frac{25}{36}\cdot\frac{36}{65}=$
$= \frac{2}{5}+\frac{5}{13}=\frac{26+25}{65}=\frac{51}{65}.$