даю 30 баллов, помогите пожалуйста. Created by ekstradiroll. algebra-ru

даю 30 баллов, помогите пожалуйста...

ekstradiroll @ekstradiroll

July 2022 1 8 Report

даю 30 баллов, помогите пожалуйста

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NNNLLL54

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

$3)\ b$

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

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

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