Question

упр 2 Вычислите b) √3/75+√3 6/25+√12,5•√2 Алгебра 9 класс

Answer 1

$\dfrac{\sqrt{3} }{\sqrt{75} }+\sqrt{3\dfrac{6}{25} }+\sqrt{12,5}\cdot\sqrt{2}=\dfrac{\sqrt{3} }{\sqrt{25\cdot3} }+\sqrt{\dfrac{81}{25} }+\sqrt{12,5\cdot2}= \\\\=\dfrac{\sqrt{3} }{5\sqrt{3} } +\dfrac{9}{5} +\sqrt{25}=\dfrac{1}{5}+\dfrac{9}{5}+5=2+5=\boxed7$

Answer 2

Ответ:  $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\dddddd\Large \boldsymbol{7}$

Объяснение:

$\displaystyle \Large \boldsymbol{} 1) \ \sqrt{a}\cdot \sqrt{b} =\sqrt{a\cdot b} \\\\\\2) \ \frac{\sqrt{a} }{\sqrt{b} }=\sqrt{\frac{a}{b} } \\\\\\ b) \ \ \frac{\sqrt{3} }{\sqrt{75} } +\sqrt{3\frac{6}{25} } +\underbrace{\sqrt{12,5} \cdot \sqrt{2} }_{\sqrt{25} }= \\\\\\\sqrt{\frac{3\!\!\!\!\diagup}{3\!\!\!\!\diagup\cdot 25} } +\sqrt{\frac{81}{25} } +5 = \\\\\\\sqrt\!\!\!\!\!\!\bigg /{\bigg(\frac{1}{5}\bigg)}^{2\!\!\! /} } +\sqrt\!\!\!\!\!\!\bigg / { \bigg(\frac{9}{5}\bigg)^{2\!\!\! /} }} +5 =$                                                            $\displaystyle\ \Large \boldsymbol{} \underbrace{ \frac{9}{5} +\frac{1}{5}}_{2} +5=2+5=\bold7$