Упрости выражения и выполни действия.

Привет! Давай упростим эти выражения и выполним действия. **1. Упростите выражение:** 1) а) $3\sqrt{c}+8\sqrt{c}-9\sqrt{c} = (3+8-9)\sqrt{c} = 2\sqrt{c}$ б) $5\sqrt{a}-2\sqrt{b}+\sqrt{a} = (5+1)\sqrt{a}-2\sqrt{b} = 6\sqrt{a}-2\sqrt{b}$ в) $\sqrt{4x}+\sqrt{64x}-\sqrt{81x} = 2\sqrt{x}+8\sqrt{x}-9\sqrt{x} = (2+8-9)\sqrt{x} = \sqrt{x}$ г) $\sqrt{27}-\sqrt{48}+\sqrt{75} = \sqrt{9\cdot3}-\sqrt{16\cdot3}+\sqrt{25\cdot3} = 3\sqrt{3}-4\sqrt{3}+5\sqrt{3} = (3-4+5)\sqrt{3} = 4\sqrt{3}$ 2) а) $\sqrt{12y}-0,5\sqrt{48y}+2\sqrt{108y} = \sqrt{4\cdot3y}-0,5\sqrt{16\cdot3y}+2\sqrt{36\cdot3y} = 2\sqrt{3y}-0,5\cdot4\sqrt{3y}+2\cdot6\sqrt{3y} = 2\sqrt{3y}-2\sqrt{3y}+12\sqrt{3y} = 12\sqrt{3y}$ б) $2\sqrt{8a}+0,3\sqrt{45c}-4\sqrt{18a}+0,01\sqrt{500c} = 2\sqrt{4\cdot2a}+0,3\sqrt{9\cdot5c}-4\sqrt{9\cdot2a}+0,01\sqrt{100\cdot5c} = 4\sqrt{2a}+0,9\sqrt{5c}-12\sqrt{2a}+0,1\sqrt{5c} = (4-12)\sqrt{2a}+(0,9+0,1)\sqrt{5c} = -8\sqrt{2a}+1\sqrt{5c} = -8\sqrt{2a}+\sqrt{5c}$ 3) а) $\sqrt{3}(\sqrt{27}-\sqrt{48}) = \sqrt{3}(\sqrt{9\cdot3}-\sqrt{16\cdot3}) = \sqrt{3}(3\sqrt{3}-4\sqrt{3}) = \sqrt{3}(-\sqrt{3}) = -3$ б) $(5\sqrt{7}-\sqrt{63}+\sqrt{14})\cdot\sqrt{7} = (5\sqrt{7}-\sqrt{9\cdot7}+\sqrt{14})\cdot\sqrt{7} = (5\sqrt{7}-3\sqrt{7}+\sqrt{14})\cdot\sqrt{7} = (2\sqrt{7}+\sqrt{14})\cdot\sqrt{7} = 2\sqrt{7}\cdot\sqrt{7}+\sqrt{14}\cdot\sqrt{7} = 2\cdot7+\sqrt{2\cdot7}\cdot\sqrt{7} = 14+\sqrt{2}\cdot\sqrt{7}\cdot\sqrt{7} = 14+7\sqrt{2}$ в) $3\sqrt{2}(2-5\sqrt{32})-2\sqrt{18} = 3\sqrt{2}(2-5\sqrt{16\cdot2})-2\sqrt{9\cdot2} = 3\sqrt{2}(2-5\cdot4\sqrt{2})-2\cdot3\sqrt{2} = 3\sqrt{2}(2-20\sqrt{2})-6\sqrt{2} = 6\sqrt{2}-60\cdot2-6\sqrt{2} = 6\sqrt{2}-120-6\sqrt{2} = -120$ г) $\sqrt{12}-(\sqrt{15}-3\sqrt{5})\sqrt{5} = \sqrt{4\cdot3}-(\sqrt{15}\cdot\sqrt{5}-3\sqrt{5}\cdot\sqrt{5}) = 2\sqrt{3}-(\sqrt{3\cdot5}\cdot\sqrt{5}-3\cdot5) = 2\sqrt{3}-(\sqrt{3}\cdot5-15) = 2\sqrt{3}-5\sqrt{3}+15 = -3\sqrt{3}+15$ **2. Выполните действия:** 1) а) $(2+\sqrt{3})(1-\sqrt{3}) = 2-2\sqrt{3}+\sqrt{3}-3 = -1-\sqrt{3}$ б) $(\sqrt{2}-\sqrt{5})(2\sqrt{2}+\sqrt{5}) = 2\cdot2+\sqrt{10}-2\sqrt{10}-5 = 4-\sqrt{10}-5 = -1-\sqrt{10}$ в) $(\sqrt{7}-\sqrt{12})(\sqrt{7}-3\sqrt{3}) = 7-3\sqrt{21}-\sqrt{84}+3\sqrt{36} = 7-3\sqrt{21}-\sqrt{4\cdot21}+3\cdot6 = 7-3\sqrt{21}-2\sqrt{21}+18 = 25-5\sqrt{21}$ г) $(2\sqrt{5}-\sqrt{18})(\sqrt{18}+\sqrt{5}) - \sqrt{200}= (2\sqrt{5}-\sqrt{9\cdot2})(\sqrt{9\cdot2}+\sqrt{5}) - \sqrt{100\cdot2}=(2\sqrt{5}-3\sqrt{2})(3\sqrt{2}+\sqrt{5}) - 10\sqrt{2}= 6\sqrt{10}+2\cdot5-9\cdot2-3\sqrt{10}-10\sqrt{2}= 6\sqrt{10}+10-18-3\sqrt{10}-10\sqrt{2}= 3\sqrt{10}-8-10\sqrt{2}$ 2) а) $(a+\sqrt{c})(a-\sqrt{c}) = a^2 - c$ б) $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) = x-y$ г) $(b+\sqrt{m})^2 = b^2+2b\sqrt{m}+m$ д) $(\sqrt{k}-\sqrt{p})^2 = k-2\sqrt{kp}+p$ в) $(\sqrt{14}-2)(2+\sqrt{14}) = (\sqrt{14}-2)(\sqrt{14}+2) = 14-4 = 10$ е) $(\sqrt{3}-\sqrt{6})^2 = 3-2\sqrt{18}+6 = 9-2\sqrt{9\cdot2} = 9-6\sqrt{2}$ 3) а) $(1+3\sqrt{2})(3\sqrt{2}-1) = 3\sqrt{2}-1+18-3\sqrt{2} = 17$ в) $(1-2\sqrt{3})^2 = 1-4\sqrt{3}+4\cdot3 = 13-4\sqrt{3}$ б) $(5\sqrt{3}-\sqrt{11})(\sqrt{11}+5\sqrt{3}) = (5\sqrt{3}-\sqrt{11})(5\sqrt{3}+\sqrt{11}) = 25\cdot3-11 = 75-11 = 64$ г) $(3\sqrt{5}+2\sqrt{10})^2 = 9\cdot5+12\sqrt{50}+4\cdot10 = 45+12\sqrt{25\cdot2}+40 = 85+12\cdot5\sqrt{2} = 85+60\sqrt{2}$ Надеюсь, теперь тебе стало понятнее!