Question

Помогите пожалуйста разобраться с 3 задачей

Answer 1

$\sqrt{1\frac{7}{9}}+10\sqrt{0,25}=\sqrt{\frac{16}{9}}+10\sqrt{\frac{1}{4}}= \sqrt{\frac{4^2}{3^2}}+10\sqrt{\frac{1}{2^2}}=\frac{4}{3}+\frac{10}{2}=1\frac{1}{3}+5=6\frac{1}{3}\\\\\\\sqrt{0,2}*\sqrt{5}-(-\sqrt{5})^2=\sqrt{0,2*5}-(\sqrt{5})^2=\sqrt{1}-5=1-5=-4$

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$\sqrt{n^2}=n\\\\\sqrt{a}*\sqrt{b}=\sqrt{ab} \\\\(\sqrt{c})^2=c$

Answer 2

$\sqrt{1\dfrac{7}{9}} + 10\sqrt{0.25} = \sqrt{\dfrac{16}{9}} + 10\sqrt{0.25} = \sqrt{\left(\dfrac{4}{3}\right)^2} + 10\sqrt{0.5^2} = \dfrac{4}{3} + 10\cdot0.5 = \dfrac{4}{3} + 5 = 6\dfrac{1}{3}$

$\sqrt{0.2}\cdot\sqrt{5} - (-\sqrt{5})^2 = \sqrt{0.2\cdot5}-(-1)^2\cdot(\sqrt{5})^2 = \sqrt{1}-(\sqrt{5})^2 = 1 - 5 = -4$