Question

Решите пожалуйстааа!

Answer 1

$1)\; \; 11\sqrt5+\sqrt{500}-\sqrt{180}=11\sqrt5+10\sqrt5-6\sqrt5=15\sqrt5\\\\3\sqrt2\cdot (5\sqrt2-\sqrt{32})=3\sqrt2\cdot (5\sqrt2-4\sqrt2)=3\sqrt2\cdot \sqrt2=6\\\\(\sqrt6+\sqrt3)^2=6+2\sqrt{6\cdot 3}+3=9+2\cdot 3\sqrt2=9+6\sqrt2\\\\(\sqrt3-\sqrt8)(\sqrt8+\sqrt3)=3-8=-5$

$2)\; \; 5\sqrt2=\sqrt{50}\; \; ,\; \; \; 4\sqrt3=\sqrt{48}\\\\Tak\; kak\; \; 50>48\; ,\; to\; \; \sqrt{50}>\sqrt{48}\; \; \Rightarrow \; \; 5\sqrt2>4\sqrt3\\\\\\3)\; \; \frac{7+\sqrt7}{5\sqrt7}=\frac{\sqrt7\cdot (\sqrt7+1)}{5\sqrt7}=\frac{\sqrt7+1}{5}\\\\\frac{16p-5}{4\sqrt{p}-\sqrt5}=\frac{(4\sqrt{p}-\sqrt5)(4\sqrt{p}+\sqrt5)}{4\sqrt{p}-\sqrt5}=4\sqrt{p}+\sqrt5$

$4)\; \; \frac{4}{2\sqrt6}=\frac{4\sqrt6}{12}\; \; ,\; \; \; \; \frac{10}{3-\sqrt5}=\frac{10\cdot (3+\sqrt5)}{9-5}=\frac{10\cdot (3+\sqrt5)}{4}=\frac{5(3+\sqrt5)}{2}\\\\\\5)\; \; \sqrt{12+6\sqrt3}=\sqrt{9+3+2\cdot 3\cdot \sqrt3}=\sqrt{(3+\sqrt3)^2}=|3+\sqrt3|=3+\sqrt3$