Question

11 класс алгебра решение пожалуйста

Answer 1

$\boxed1 \\\ \\ \left (\dfrac{3x^{\frac{1}{2}}}{3-x^{\frac{1}{2}}}+3 \right ) (9-6x^{\frac{1}{2}}+x )= \dfrac{3x^{\frac{1}{2}}+9-3x^{\frac{1}{2}}}{3-x^{\frac{1}{2}}}(3-x^{\frac{1}{2}})^2=\\ \\= 9(3-x^{\frac{1}{2}})=27-9x^{\frac{1}{2}}=27-117=-90\\ \\ \boxed2 \\ \\ 7 \left(\dfrac{a-16b}{\sqrt{a}-4\sqrt{b}} -\dfrac{a\sqrt{a}-64 b \sqrt{b}}{a-16b} \right)= \\ =7 \left(\dfrac{(\sqrt{a}-4\sqrt{b})(\sqrt{a}+4\sqrt{b})}{\sqrt{a}-4\sqrt{b}} -\dfrac{a\sqrt{a}-64 b \sqrt{b}}{a-16b} \right)=$

$=7 \dfrac{(\sqrt{a}+4\sqrt{b})(a-16b)-a\sqrt{a}+64b\sqrt{b}}{a-16b} =\\ =7\dfrac{a\sqrt{a}-16b\sqrt{a}+4a\sqrt{b}-64b\sqrt{b}-a\sqrt{a}+64b\sqrt{b}}{a-16b} =\\ = 7 \dfrac{4a\sqrt{b}-16b\sqrt{a}}{a-16b} =\\ =7\dfrac{4\sqrt{ab}}{\sqrt{a}+4\sqrt{b}} =\\ =7\dfrac{4\sqrt{4 \cdot 0,04}}{\sqrt{4}+4\sqrt{0,04}} =\\ =7\dfrac{4\cdot2\cdot0,2}{2+0,8}= 4$