Question

Срочно!!!!!!Даю 50баллов

Answer 1

$1)\; \; 9<13<16\; \; \to \; \; \sqrt9<\sqrt{13}<\sqrt{16}\; \; \to \; \; \; 3<\sqrt{13}<4\\\\144<167<169\; \; \to \; \; \sqrt{144}<\sqrt{167}<\sqrt{169}\; \; \to \; \; 12<\sqrt{167}<13\\\\\\2)\; \; 12\sqrt{400}-19\sqrt{144}+0,33=12\cdot 20-19\cdot 12+0,33=12,33\\\\\\3)\; \; a)\; \; 3\sqrt9=3\cdot 3=9\; \; ,\; \; \sqrt{729}=27\; \; \to \; \; \; 9<27\; \; \to \; \; 3\sqrt9<\sqrt{729}\\\\b)\; \; \frac{3}{4}\sqrt{32}=\frac{3}{4}\cdot \sqrt{2^5}=\frac{3}{4}\cdot 2^2\cdot \sqrt2=3\sqrt2$

$\frac{2}{7}\sqrt{98}=\frac{2}{7}\cdot \sqrt{7^2\cdot 2}=\frac{2}{7}\cdot 7\cdot \sqrt2=2\sqrt2\\\\3\sqrt2>2\sqrt2\; \; \to \; \; \; \frac{3}{4}\sqrt{32}>\frac{2}{7}\sqrt{98}\\\\\\4)\; \; \sqrt{75}-2\sqrt3\cdot (3-8\sqrt{12})=\sqrt{25\cdot 3}-2\sqrt3\cdot (3-8\sqrt{4\cdot 3})=\\\\=5\sqrt3-2\sqrt3\cdot (3-8\cdot 2\sqrt3)=5\sqrt3-6\sqrt3+32\cdot 3=96-\sqrt3$

$5a)\; \; (2\sqrt{11}-\sqrt{11a})(2\sqrt{11}+\sqrt{11a})=(2\sqrt{11})^2-(\sqrt{11a})^2=\\\\=4\cdot 11-11a=11\cdot (4-a)\\\\\\5b)\; \; \Big (\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}}\Big )\cdot \frac{x-y}{x^2+xy}=\frac{\sqrt{x}(\sqrt{x}+\sqrt{y})-\sqrt{y}(\sqrt{x}-\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\cdot \frac{x-y}{x(x+y)}=\\\\=\frac{x+y}{x-y}\cdot \frac{x-y}{x(x+y)}=\frac{1}{x}$

Answer 2

Ответ во вложении Объяснение: