Question

17-√17/√17=
30/7√5=
35/√37+√2=​

Answer 1

Ответ:

$\boxed{\ a=\sqrt{a}\cdot \sqrt{a}=(\sqrt{a})^2\ \ ,\ \ (a-b)(a+b)=a^2-b^2\ \ }$

$\displaystyle 1)\ \frac{17-\sqrt{17}}{\sqrt{17}}=\frac{\sqrt{17}\cdot (\sqrt{17}-1)}{\sqrt{17}}=\sqrt{17}-1\\\\\\2)\ \frac{30}{7\sqrt5}=\frac{6\cdot \sqrt{5}\cdot \sqrt{5}}{7\sqrt5}=\dfrac{6\sqrt5}{7}\\\\\\3)\ \frac{35}{\sqrt{37}+\sqrt2}=\frac{35(\sqrt{37}-\sqrt2)}{(\sqrt{37}+\sqrt2)(\sqrt{37}-\sqrt2)}}=\frac{35(\sqrt{37}-\sqrt2)}{37-2}=\frac{35(\sqrt{37}-\sqrt2)}{35}=\\\\\\=\sqrt{37}-\sqrt2$