Question

ДАМ 100 БАЛЛОВ ВАРИАНТ 8.7.2 ОТ 1-4

Answer 1

Ответ:

$1)\ \ 2\sqrt3-\sqrt{75}=2\sqrt3-\sqrt{25\cdot 3}=2\sqrt3-5\sqrt3=-3\sqrt3\\\\2)\ \ \sqrt{18}+\sqrt{50}=\sqrt{9\cdot 2}+\sqrt{25\cdot 2}=3\sqrt2+5\sqrt2=8\sqrt2\\\\3)\ \ x>0\ ,\ \ \dfrac{4\sqrt{x^2y}}{2x\sqrt{y}}=\dfrac{4\, |x|\sqrt{y}}{2x\sqrt{y}}=\dfrac{4x\sqrt{y}}{2x\sqrt{y}}=2\\\\4)\ \ y>0\ ,\ \ \dfrac{\sqrt5}{y}+\sqrt{\dfrac{20}{y^2}}=\dfrac{\sqrt5}{y}+\dfrac{\sqrt{20}}{|y|}=\dfrac{\sqrt5}{y}+\dfrac{2\sqrt{5}}{y}=\dfrac{\sqrt5+2\sqrt5}{y}=\dfrac{3\sqrt5}{y}$

$5)\ \ (\sqrt3+2)^2=3+4\sqrt3+4=7+4\sqrt3\\\\6)\ \ (\sqrt3-\sqrt{12})^2=3-2\sqrt{3\cdot 12}+12=15-2\sqrt{3\cdot 3\cdot 4}=15-2\cdot 3\cdot 2=15-12=3\\\\7)\ \ \sqrt{(\sqrt5-2)^2}=|\underbrace{\sqrt5-2}_{>0}|=\sqrt5-2\\\\8)\ \ \sqrt{(\sqrt{5}-3)^2}=|\underbrace{\sqrt5-3}_{<0}|=-(\sqrt5-3)=3-\sqrt5$

$9)\ \ \dfrac{\sqrt{\sqrt{13}+2}}{\sqrt{\sqrt{13}-2}}=\dfrac{\sqrt{(\sqrt{13}+2)^2}}{\sqrt{(\sqrt{13}-2)(\sqrt{13}+2)}}=\dfrac{|\sqrt{13}+2|}{\sqrt{13-4}}=\dfrac{\sqrt{13}+2}{\sqrt9}=\dfrac{\sqrt{13}+2}{3}$