№ 16.15 Алгебра = № 32.15 Математика
Обчисліть:
1. $\left(-(\sqrt5)^2\right)^2;$
2. $(2\sqrt5)^2-(5\sqrt2)^2;$
3. $36\cdot\left(-\frac{1}{3}\sqrt{17}\right)^2-\frac{1}{5}(2\sqrt{15})^2;$
4. $\sqrt{59{,}29}+\left(\frac{1}{2}\sqrt{34}\right)^2;$
5. $(-3\sqrt5)^2-3(\sqrt5)^2;$
6. $\left(-\frac{4}{5}\cdot\sqrt{\frac{25}{32}}\right)^2-\left(\frac{3}{4}\cdot\sqrt{\frac{8}{9}}\right)^2.$
Розв'язок:
1. $\left(-(\sqrt5)^2\right)^2=\left((\sqrt5)^2\right)^2=$
$= 5^2=25;$
2. $(2\sqrt5)^2-(5\sqrt2)^2=$
$= 4·5-25·2=20-50=$
$= -30;$
3. $36\cdot\left(-\frac{1}{3}\sqrt{17}\right)^2-$
$- \frac{1}{5}(2\sqrt{15})^2=$
$= 36\cdot\frac{1}{9}\cdot17-\frac{1}{5}\cdot4\cdot15=$
$= 4·17-4·3=$
$= 4·(17-3)= 4·14=56;$
4. $\sqrt{59{,}29}+\left(\frac{1}{2}\sqrt{34}\right)^2=$
$= 7{,}7+\frac{1}{4}\cdot34=7{,}7+\frac{17}{2}=$
$= 7{,}7+8{,}5= 16{,}2;$
5. $(-3\sqrt5)^2-3(\sqrt5)^2=$
$= 9\cdot5-3\cdot5 =5\cdot(9-3)=$
$= 5\cdot6=30;$
6. $\left(-\frac{4}{5}\cdot\sqrt{\frac{25}{32}}\right)^2-$
$- \left(\frac{3}{4}\cdot\sqrt{\frac{8}{9}}\right)^2=$
$=\frac{16}{25}\cdot\frac{25}{32}-\frac{9}{16}\cdot\frac{8}{9}=$
$= \frac{1}{2}-\frac{1}{2}=0.$