Question

ПОМОГИТЕ РЕШИТЬ ЧЕТЫРЕ ПРИМЕРА!!!СРОЧНО ПОЖАЛУЙСТА!!!

Answer 1

$<var>\frac{9x-y}{3x+x^{0,5}y^{0,5}}=\frac{(3x^{0,5}+y^{0,5})(3x^{0,5}-y^{0,5})}{x^{0,5}(3x^{0,5}+y^{0,5})}=\frac{3x^{0,5}-y^{0,5}}{x^{0,5}}=3-\frac{y^{0,5}}{x^{0,5}}</var>$

$<var>3-\frac{576^{0,5}}{100^{0,5}}=3-\frac{24}{10}=3-2,4=0,6</var>$

 

$<var>40^{\frac{1}{3}}+162^{\frac{1}{4}}-3\sqrt[4]2-2\sqrt[3]5=40^{\frac{1}{3}}+162^{\frac{1}{4}}-\sqrt[4]{2*3^4}-\sqrt[3]{5*2^3}=\\\ =40^{\frac{1}{3}}+162^{\frac{1}{4}}-\sqrt[4]{162}-\sqrt[3]{40}=40^{\frac{1}{3}}+162^{\frac{1}{4}}-162^{\frac{1}{4}}-40^{\frac{1}{3}}=0</var>$

 

$<var>\sqrt{10-\sqrt{96}}-\sqrt{10+\sqrt{96}}=\sqrt{10-4\sqrt{6}}-\sqrt{10+4\sqrt{6}}=\\\ =\sqrt{2^2-2*2*\sqrt{6}+(\sqrt{6})^2}-\sqrt{2^2+2*2*\sqrt{6}+(\sqrt{6})^2}}=\\\ =\sqrt{(2-\sqrt{6})^2}-\sqrt{(2+\sqrt{6})^2}=|2-\sqrt{6}|-|2+\sqrt{6}|=\\\ =\sqrt{6}-2-2-\sqrt{6}=-4</var>$

 

$<var>\frac{7\sqrt{30}}{3\sqrt{10}-10\sqrt{3}}+\sqrt{3}+\sqrt{10}=\frac{7\sqrt{30}}{\sqrt{3}\sqrt{10}(\sqrt{3}-\sqrt{10})}+\sqrt{3}+\sqrt{10}=\\\ \frac{7}{\sqrt{3}-\sqrt{10}}+\sqrt{3}+\sqrt{10}=\frac{7+(\sqrt{3}+\sqrt{10})(\sqrt{3}-\sqrt{10})}{\sqrt{3}-\sqrt{10}}=\frac{7+(3-10)}{\sqrt{3}-\sqrt{10}}=0</var>$