Question

Помогите пожалуйста срочно!!! С объяснением!!! Алгебра 8 класс!!!

Answer 1

$1)\; \; (3\sqrt{6}+5\sqrt{8}-4\sqrt{32})\cdot \sqrt2-\sqrt{108}=3\sqrt{12}+5\sqrt{16}-4\sqrt{64}-\sqrt{4\cdot 27}=\\\\=3\cdot 2\sqrt3+5\cdot 4-4\cdot 8-2\cdot 3\sqrt3=20-32=-12\\\\\\2)\; \; (\sqrt5+7\sqrt2)(7\sqrt2-\sqrt5)-(\sqrt{10}-2\sqrt5)^2=\\\\=(7\sqrt2)^2-(\sqrt5)^2-((\sqrt{10})^2-2\cdot \sqrt{10}\cdot 2\sqrt5+(2\sqrt5)^2)=\\\\=49\cdot 2-5-10+4\sqrt{50}-4\cdot 5=63+4\cdot 5\sqrt2=63+20\sqrt2$

$3)\; \; (7-\sqrt3)^2+(4+\sqrt3)^2=49-14\sqrt3+3+16+8\sqrt3+3=71-6\sqrt3\\\\\\4)\; \; (\; \sqrt{7-4\sqrt3}+\sqrt{7+4\sqrt3}\; )^2=7-4\sqrt3+7+4\sqrt3+2\sqrt{(7-4\sqrt3)(7+4\sqrt3)}=\\\\=14+2\sqrt{49-48}=14+2\cdot 1=16$

$5)\; \; \frac{x^2-11}{x+\sqrt{11}}=\frac{(x-\sqrt{11})(x+\sqrt{11})}{x+\sqrt{11}}=x-\sqrt{11}\\\\6)\; \; \frac{\sqrt{x}-12}{x-144}=\frac{\sqrt{x}-12}{(\sqrt{x}-12)(\sqrt{x}+12)}=\frac{1}{\sqrt{x}+12}\\\\7)\; \; \frac{a+3\sqrt{a}}{a-9}=\frac{\sqrt{a}\, (\sqrt{a}+3)}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{\sqrt{a}}{\sqrt{a}-3}\\\\8)\; \; \frac{17-\sqrt{17}}{\sqrt{17}}=\frac{\sqrt{17}\, (\sqrt{17}-1)}{\sqrt{17}}=\sqrt{17}-1$

$9)\; \; \frac{m-12\sqrt{m}+36}{m-36}=\frac{(\sqrt{m}-6)^2}{(\sqrt{m}-6)(\sqrt{m}+6)}=\frac{\sqrt{m}-6}{\sqrt{m}+6}\\\\10)\; \; \frac{\sqrt{21}-3}{7-\sqrt{21}}=\frac{\sqrt3\, (\sqrt7-\sqrt3)}{\sqrt7\, (\sqrt7-\sqrt3)}=\frac{\sqrt3}{\sqrt7}=\sqrt{\frac{3}{7}}$