Question

СРОЧНО!!! АЛГЕБРА 8 КЛАСС

Answer 1

A1.

a) $\sqrt{900} = 30$;

б) $\sqrt{3 \frac{1}{16}} = \sqrt{\frac{49}{16}} = \frac{7}{4}$.

A2.

a) $\sqrt{225 \cdot 0,09} = \sqrt{20.25} = 4.5$;

б) $\frac{\sqrt{108}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$;

в) $\sqrt{6^{4}} = \sqrt{1296} = 36$.

A3.

$\sqrt{49 c}-\sqrt{16 c}+\sqrt{25 c} = 7\sqrt{c} - 4\sqrt{c} + 5\sqrt{c} = 8\sqrt{c}$.

A4.

a) $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$;

б) $\frac{6}{\sqrt{5}+1} = \frac{6(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} = \frac{6(\sqrt{5}-1)}{5-1} = \frac{3(\sqrt{5}-1)}{2}$.

A. В.

a) $\sqrt{117} = \sqrt{9 \cdot 13} = 3\sqrt{13}$;

б) $\sqrt{5 c^{2}}$, де $c < 0$ - за умовою $c < 0$, отже $\sqrt{5 c^{2}} = \sqrt{5} \cdot |c|$.

B1.

a) $(3 \sqrt{7}+4)(3 \sqrt{7}-4) = (3\sqrt{7})^2 - 4^2 = 63 - 16 = 47$;

б) $(4 \sqrt{5}-\sqrt{3})^{2}+8 \sqrt{15} = (4\sqrt{5})^2 - 2 \cdot 4 \sqrt{5} \cdot \sqrt{3} + (\sqrt{3})^2 + 8\sqrt{15} = 80 - 24 + 3 + 8\sqrt{15} = 59 + 8\sqrt{15}$.

B2.

$\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10}-\sqrt{6}}-\frac{\sqrt{10}-\sqrt{6}}{\sqrt{10}+\sqrt{6}} = \frac{(\sqrt{10}+\sqrt{6})^2 - (\sqrt{10}-\sqrt{6})^2}{(\sqrt{10}-\sqrt{6})(\sqrt{10}+\sqrt{6})}$
$= \frac{10 + 2\sqrt{60} + 6 - (10 - 2\sqrt{60} + 6)}{10 - 6} = \frac{4\sqrt{60}}{4} = \sqrt{60}$