Question

даю 30 баллов, помогите пожалуйста

Answer 1

Ответ:

$\displaystyle 1)\ \ \dfrac{2}{5}\sqrt{62\dfrac{1}{2}}=\sqrt{\frac{4}{25}\cdot\frac{125}{2}}=\sqrt{\frac{2}{1}\cdot \frac{5}{1}}=\sqrt{10}\\\\\\\frac{4}{3}\sqrt{5\frac{5}{8}}=\sqrt{\frac{16}{9}\cdot \frac{45}{8}}=\sqrt{10}\\\\\\\dfrac{2}{5}\sqrt{62\dfrac{1}{2}}=\frac{4}{3}\sqrt{5\frac{5}{8}}\\\\\\2)\ \ a\geq 0\ \ ,\ \ \sqrt{3a^2}=|a|\cdot \sqrt3=a\sqrt3$

$3)\ b<0\ \ ,\ \ \sqrt{242m^{11}b^{18}}=\sqrt{11^2\cdot 2\cdot m^{10}\cdot m\cdot (b^9)^2}=11\cdot |\underbrace{m^5}_{\geq 0}|\cdot |\underbrace{b^9}_{<0}|\cdot \sqrt{2m}=\\\\=11m^5\cdot (-b^9)\, \sqrt{2m}=-11\, m^5\, b^9\, \sqrt{2m}\ ,\\\\(\ tak\ kak\ \ 242m^{11}b^{18}\geq 0\ ,\ to \ \ m^{11}\geq 0\ \ \to \ \ m\geq 0\ )$

$\displaystyle 4)\ \ a\geq 0\ \ ,\ \ ab^2\sqrt{\frac{a}{b}}=\sqrt{a^2b^3\cdot \frac{a}{b}}=\sqrt{a^3b^3}\\\\\\5)\ \ 2\sqrt{10}=\sqrt{4\cdot 10}=\sqrt{40}\\\\38<40\ \ \ \Rightarrow \ \ \ \sqrt{38}<\sqrt{40}\ \ \ \Rightarrow \ \ \ \ \sqrt{38}<2\sqrt{10}$

$6)\ \ \sqrt{450}=\sqrt{225\cdot 2}=\sqrt{15^2\cdot 2}=15\sqrt2\\\\\\7)\ \ -\dfrac{2}{3}\sqrt{54}=-\sqrt{\dfrac{4\cdot 54}{9} }=-\sqrt{4\cdot 6}=-\sqrt{24}\\\\\\8)\ \ \dfrac{2}{9}\sqrt{\dfrac{27}{28}}=\sqrt{\dfrac{4\cdot 27}{81\cdot 28} }=\sqrt{\dfrac{1}{3\cdot 7}}=\sqrt{\dfrac{1}{21}}$