Алгебра: 1.Решите 1.√18*√2 - √20 * √5 = 2.(√20 - √80) √5 = 3.(2√7 +3)Квадрат = 2....

Решите задачу:

$1)\; \; \sqrt{18}\cdot \sqrt{2}-\sqrt{20}\cdot \sqrt{5}=\sqrt{18\cdot 2}-\sqrt{20\cdot 5} = \sqrt{36}-\sqrt{100}=\\\\=6-10=-4\\\\2)\; \; ( \sqrt{20}-\sqrt{80})\cdot \sqrt{5}=\sqrt{100}-\sqrt{400}=10-20=-10\\\\3)\; \; (2\sqrt{7}+3)^2=4\cdot 7+12\sqrt{7}+9=37+12\sqrt{7}\\\\4)\; \; 2\sqrt{18}-0,5 \sqrt{8}+\sqrt{50}=2\cdot 3\sqrt{2}-0,5\cdot 2\sqrt2+5\sqrt2=\\\\=\sqrt2\cdot (6-1+5)=\sqrt2\cdot 10=10\sqrt2\\\\5)\; \; (\sqrt5+\sqrt3)(\sqrt5-\sqrt3)=(\sqrt5)^2-(\sqrt3)^2=5-3=2$

$6)\; \; 2\sqrt6=\sqrt{2^2\cdot 6}=\sqrt{24}\\\\4\sqrt2=\sqrt{4^2\cdot 2}=\sqrt{32}\\\\24\ \textless \ 32\; \; \; \Rightarrow \; \; \; \sqrt{24}\ \textless \ \sqrt{32}\; \; \; \Rightarrow \; \; \; 2\sqrt6\ \textless \ 4\sqrt2$

$7)\; \; \frac{\sqrt5+5}{4\sqrt5}=\frac{\sqrt5(1+\sqrt5))}{4\sqrt5}=\frac{1+\sqrt5}{4} \\\\8)\; \; \frac{25-x}{\sqrt{x}+5}=\frac{(5- \sqrt{x})(5+\sqrt{x})}{\sqrt{x}+5}=5- \sqrt{x}$