Сократить дробь с корнем :. Created by Maria89036. matematika-ru

Сократить дробь с корнем :...

Петросян4

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Ответ с вложением, а то неудобно на телефоне писать.

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skvrttt

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а) $\frac{a^2-2}{a-\sqrt{2}}=\frac{(a-\sqrt{2})(a+\sqrt{2})}{a-\sqrt{2}}=a+\sqrt{2}$ ;
б) $\frac{\sqrt{13}+13}{\sqrt{13}}=\frac{\sqrt{13}}{\sqrt{13}}+\frac{13}{\sqrt{13}}=1+\sqrt{13}$ ;
в) $\frac{a^4-3}{a^2+\sqrt{3}}=\frac{(a^2-\sqrt{3})(a^2+\sqrt{3})}{a^2+\sqrt{3}}=a^2-\sqrt{3}$ ;
г) $\frac{3\sqrt{2}-2}{6\sqrt{2}}=\frac{3\sqrt{2}}{6\sqrt{2}}-\frac{2}{6\sqrt{2}}=\frac{1}{2}-\frac{1}{3\sqrt{2}}=0,5-\frac{\sqrt{2}}{6}$ ;
д) $\frac{\sqrt{c}-3}{9-c}=\frac{\sqrt{c}-3}{(3-\sqrt{c})(3+\sqrt{c})}=-\frac{1}{3+\sqrt{c}}=-(3+\sqrt{c})^{-1}$ ;
е) $\frac{3+\sqrt{6}}{\sqrt{6}}=\frac{3}{\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}}{2}+1$ .

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