Giúp mình với

Bài 11: a) $\sqrt{20}+\sqrt{45}+\sqrt{80}$ Ta có: $\sqrt{20}=\sqrt{4 \cdot 5}=2\sqrt{5}$ $\sqrt{45}=\sqrt{9 \cdot 5}=3\sqrt{5}$ $\sqrt{80}=\sqrt{16 \cdot 5}=4\sqrt{5}$ Do đó: $\sqrt{20}+\sqrt{45}+\sqrt{80}=2\sqrt{5}+3\sqrt{5}+4\sqrt{5}=9\sqrt{5}$ b) $2\sqrt{50}-3\sqrt{18}+\sqrt{32}$ Ta có: $2\sqrt{50}=2\sqrt{25 \cdot 2}=2 \cdot 5\sqrt{2}=10\sqrt{2}$ $3\sqrt{18}=3\sqrt{9 \cdot 2}=3 \cdot 3\sqrt{2}=9\sqrt{2}$ $\sqrt{32}=\sqrt{16 \cdot 2}=4\sqrt{2}$ Do đó: $2\sqrt{50}-3\sqrt{18}+\sqrt{32}=10\sqrt{2}-9\sqrt{2}+4\sqrt{2}=5\sqrt{2}$ c) $\frac{1}{2}\sqrt{12}+\sqrt{27}-\frac{3}{4}\sqrt{48}$ Ta có: $\frac{1}{2}\sqrt{12}=\frac{1}{2}\sqrt{4 \cdot 3}=\frac{1}{2} \cdot 2\sqrt{3}=\sqrt{3}$ $\sqrt{27}=\sqrt{9 \cdot 3}=3\sqrt{3}$ $\frac{3}{4}\sqrt{48}=\frac{3}{4}\sqrt{16 \cdot 3}=\frac{3}{4} \cdot 4\sqrt{3}=3\sqrt{3}$ Do đó: $\frac{1}{2}\sqrt{12}+\sqrt{27}-\frac{3}{4}\sqrt{48}=\sqrt{3}+3\sqrt{3}-3\sqrt{3}=\sqrt{3}$ d) $\sqrt{(2-\sqrt{3})^2}+\sqrt{(4-\sqrt{3})^2}$ Ta có: $\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}$ (vì $2>\sqrt{3}$) $\sqrt{(4-\sqrt{3})^2}=|4-\sqrt{3}|=4-\sqrt{3}$ (vì $4>\sqrt{3}$) Do đó: $\sqrt{(2-\sqrt{3})^2}+\sqrt{(4-\sqrt{3})^2}=2-\sqrt{3}+4-\sqrt{3}=6-2\sqrt{3}$ e) $\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}$ Ta có: $\sqrt{5-2\sqrt{6}}=\sqrt{(\sqrt{3}-\sqrt{2})^2}=\sqrt{3}-\sqrt{2}$ $\sqrt{5+2\sqrt{6}}=\sqrt{(\sqrt{3}+\sqrt{2})^2}=\sqrt{3}+\sqrt{2}$ Do đó: $\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=(\sqrt{3}-\sqrt{2})-(\sqrt{3}+\sqrt{2})=-2\sqrt{2}$ g) $\sqrt{14+8\sqrt{3}}-\sqrt{14-8\sqrt{3}}$ Ta có: $\sqrt{14+8\sqrt{3}}=\sqrt{(\sqrt{6}+2\sqrt{2})^2}=\sqrt{6}+2\sqrt{2}$ $\sqrt{14-8\sqrt{3}}=\sqrt{(\sqrt{6}-2\sqrt{2})^2}=\sqrt{6}-2\sqrt{2}$ Do đó: $\sqrt{14+8\sqrt{3}}-\sqrt{14-8\sqrt{3}}=(\sqrt{6}+2\sqrt{2})-(\sqrt{6}-2\sqrt{2})=4\sqrt{2}$ h) $\sqrt[3]{-64}+(\sqrt[3]{-9})^3-\sqrt[3]{(-11)^3}$ Ta có: $\sqrt[3]{-64}=-4$ $(\sqrt[3]{-9})^3=-9$ $\sqrt[3]{(-11)^3}=-11$ Do đó: $\sqrt[3]{-64}+(\sqrt[3]{-9})^3-\sqrt[3]{(-11)^3}=-4-9+11=-2$ k) $\sqrt[3]{27}-\sqrt[3]{-8}+\sqrt[3]{-125}$ Ta có: $\sqrt[3]{27}=3$ $\sqrt[3]{-8}=-2$ $\sqrt[3]{-125}=-5$ Do đó: $\sqrt[3]{27}-\sqrt[3]{-8}+\sqrt[3]{-125}=3+2-5=0$ m) $\sqrt[3]{-64}-\sqrt[3]{-216}+\sqrt[3]{729}$ Ta có: $\sqrt[3]{-64}=-4$ $\sqrt[3]{-216}=-6$ $\sqrt[3]{729}=9$ Do đó: $\sqrt[3]{-64}-\sqrt[3]{-216}+\sqrt[3]{729}=-4+6+9=11$ n) $2\sqrt{2}+\sqrt{(\sqrt{8}-3)^2}$ Ta có: $\sqrt{(\sqrt{8}-3)^2}=|\sqrt{8}-3|=3-\sqrt{8}$ (vì $\sqrt{8}< 3$) Do đó: $2\sqrt{2}+\sqrt{(\sqrt{8}-3)^2}=2\sqrt{2}+3-\sqrt{8}=2\sqrt{2}+3-2\sqrt{2}=3$ o) $(\sqrt{48}-\sqrt{75}+6\sqrt{3}):\sqrt{3}$ Ta có: $\sqrt{48}=\sqrt{16 \cdot 3}=4\sqrt{3}$ $\sqrt{75}=\sqrt{25 \cdot 3}=5\sqrt{3}$ Do đó: $(\sqrt{48}-\sqrt{75}+6\sqrt{3}):\sqrt{3}=(4\sqrt{3}-5\sqrt{3}+6\sqrt{3}):\sqrt{3}=5\sqrt{3}:\sqrt{3}=5$ p) $2\sqrt{5}-3\sqrt{45}+\sqrt{500}$ Ta có: $3\sqrt{45}=3\sqrt{9 \cdot 5}=3 \cdot 3\sqrt{5}=9\sqrt{5}$ $\sqrt{500}=\sqrt{100 \cdot 5}=10\sqrt{5}$ Do đó: $2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}$ r) $3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}$ Ta có: $4\sqrt{18}=4\sqrt{9 \cdot 2}=4 \cdot 3\sqrt{2}=12\sqrt{2}$ $2\sqrt{32}=2\sqrt{16 \cdot 2}=2 \cdot 4\sqrt{2}=8\sqrt{2}$ $\sqrt{50}=\sqrt{25 \cdot 2}=5\sqrt{2}$ Do đó: $3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}$ s) $4\sqrt{12}+\sqrt{108}-8\sqrt{3}+\sqrt{7-4\sqrt{3}}$ Ta có: $4\sqrt{12}=4\sqrt{4 \cdot 3}=4 \cdot 2\sqrt{3}=8\sqrt{3}$ $\sqrt{108}=\sqrt{36 \cdot 3}=6\sqrt{3}$ $\sqrt{7-4\sqrt{3}}=\sqrt{(\sqrt{3}-2)^2}=\sqrt{3}-2$ Do đó: $4\sqrt{12}+\sqrt{108}-8\sqrt{3}+\sqrt{7-4\sqrt{3}}=8\sqrt{3}+6\sqrt{3}-8\sqrt{3}+\sqrt{3}-2=7\sqrt{3}-2$ Bài 12: a) Ta có $\sqrt{\frac8{13}}=\frac{\sqrt8}{\sqrt{13}}=\frac{\sqrt8\times \sqrt{13}}{\sqrt{13}\times \sqrt{13}}=\frac{\sqrt{104}}{13}.$ b) Ta có $\sqrt{\frac{-5}{-18}}=\sqrt{\frac5{18}}=\frac{\sqrt5}{\sqrt{18}}=\frac{\sqrt5\times \sqrt{18}}{\sqrt{18}\times \sqrt{18}}=\frac{\sqrt{90}}{18}.$ c) Ta có $\sqrt{\frac6{25}}=\frac{\sqrt6}{\sqrt{25}}=\frac{\sqrt6}{5}.$ d) Ta có $\sqrt{\frac{8x^2}y}=\frac{x\sqrt8}{\sqrt{y}}=\frac{x\sqrt8\times \sqrt{y}}{\sqrt{y}\times \sqrt{y}}=\frac{x\sqrt{8y}}{y}.$ Bài 13: a) $\frac{1}{\sqrt{7}+2}$ Nhân tử và mẫu với $\sqrt{7}-2$: $\frac{1}{\sqrt{7}+2} \cdot \frac{\sqrt{7}-2}{\sqrt{7}-2} = \frac{\sqrt{7}-2}{(\sqrt{7})^2-(2)^2} = \frac{\sqrt{7}-2}{7-4} = \frac{\sqrt{7}-2}{3}$ b) $\frac{6}{\sqrt{7}-\sqrt{3}}$ Nhân tử và mẫu với $\sqrt{7}+\sqrt{3}$: $\frac{6}{\sqrt{7}-\sqrt{3}} \cdot \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} = \frac{6(\sqrt{7}+\sqrt{3})}{(\sqrt{7})^2-(\sqrt{3})^2} = \frac{6(\sqrt{7}+\sqrt{3})}{7-3} = \frac{6(\sqrt{7}+\sqrt{3})}{4} = \frac{3(\sqrt{7}+\sqrt{3})}{2}$ c) $\frac{5}{2\sqrt{3}-\sqrt{2}}$ Nhân tử và mẫu với $2\sqrt{3}+\sqrt{2}$: $\frac{5}{2\sqrt{3}-\sqrt{2}} \cdot \frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{3}+\sqrt{2}} = \frac{5(2\sqrt{3}+\sqrt{2})}{(2\sqrt{3})^2-(\sqrt{2})^2} = \frac{5(2\sqrt{3}+\sqrt{2})}{12-2} = \frac{5(2\sqrt{3}+\sqrt{2})}{10} = \frac{2\sqrt{3}+\sqrt{2}}{2}$ d) $\frac{-4}{4+3\sqrt{2}}$ Nhân tử và mẫu với $4-3\sqrt{2}$: $\frac{-4}{4+3\sqrt{2}} \cdot \frac{4-3\sqrt{2}}{4-3\sqrt{2}} = \frac{-4(4-3\sqrt{2})}{(4)^2-(3\sqrt{2})^2} = \frac{-4(4-3\sqrt{2})}{16-18} = \frac{-4(4-3\sqrt{2})}{-2} = 2(4-3\sqrt{2}) = 8-6\sqrt{2}$ k) $\frac{a-16}{2\sqrt{a}-8}$ với $a \geq 0$ Nhân tử và mẫu với $2\sqrt{a}+8$: $\frac{a-16}{2\sqrt{a}-8} \cdot \frac{2\sqrt{a}+8}{2\sqrt{a}+8} = \frac{(a-16)(2\sqrt{a}+8)}{(2\sqrt{a})^2-(8)^2} = \frac{(a-16)(2\sqrt{a}+8)}{4a-64} = \frac{(a-16)(2\sqrt{a}+8)}{4(a-16)} = \frac{2\sqrt{a}+8}{4} = \frac{\sqrt{a}+4}{2}$ m) $\frac{4a-1}{4\sqrt{a}+2}$ với $a \geq 0$ Nhân tử và mẫu với $4\sqrt{a}-2$: $\frac{4a-1}{4\sqrt{a}+2} \cdot \frac{4\sqrt{a}-2}{4\sqrt{a}-2} = \frac{(4a-1)(4\sqrt{a}-2)}{(4\sqrt{a})^2-(2)^2} = \frac{(4a-1)(4\sqrt{a}-2)}{16a-4} = \frac{(4a-1)(4\sqrt{a}-2)}{4(4a-1)} = \frac{4\sqrt{a}-2}{4} = \frac{2\sqrt{a}-1}{2}$ Bài 14: a) $A=(\frac{\sqrt x+1}{\sqrt x-1}-\frac{\sqrt x-1}{\sqrt x+1}).\frac{x-1}4$ $= [\frac{(\sqrt x+1)^2-(\sqrt x-1)^2}{(\sqrt x-1)(\sqrt x+1)}].\frac{x-1}4$ $= [\frac{(x+2\sqrt x+1)-(x-2\sqrt x+1)}{(\sqrt x-1)(\sqrt x+1)}].\frac{x-1}4$ $= [\frac{4\sqrt x}{(\sqrt x-1)(\sqrt x+1)}].\frac{x-1}4$ $= \frac{4\sqrt x(x-1)}{4(\sqrt x-1)(\sqrt x+1)}$ $= \frac{\sqrt x(x-1)}{(\sqrt x-1)(\sqrt x+1)}$ $= \frac{\sqrt x(\sqrt x-1)(\sqrt x+1)}{(\sqrt x-1)(\sqrt x+1)}$ $= \sqrt x$ b) $B=\frac{a-16}{\sqrt a-4}+\frac{a-9}{\sqrt a-3}$ $= \frac{(\sqrt a-4)(\sqrt a+4)}{\sqrt a-4}+\frac{(\sqrt a-3)(\sqrt a+3)}{\sqrt a-3}$ $= \sqrt a+4+\sqrt a+3$ $= 2\sqrt a+7$ c) $C=\frac{a-10\sqrt a+25}{\sqrt a-5}-\frac{a+3\sqrt a}{\sqrt a+3}$ $= \frac{(\sqrt a-5)^2}{\sqrt a-5}-\frac{\sqrt a(\sqrt a+3)}{\sqrt a+3}$ $= \sqrt a-5-\sqrt a$ $= -5$ d) $D=(\frac{a-4}{\sqrt a+2}+\frac{2a-8}{\sqrt a-2}).(3\sqrt a-2)$ $= (\frac{(\sqrt a-2)(\sqrt a+2)}{\sqrt a+2}+\frac{2(\sqrt a-2)(\sqrt a+2)}{\sqrt a-2}).(3\sqrt a-2)$ $= (\sqrt a-2+2(\sqrt a+2)).(3\sqrt a-2)$ $= (3\sqrt a+2).(3\sqrt a-2)$ $= 9a-4$ e) $E=\frac{x-2\sqrt x+1}{\sqrt x-2}-\frac{16-x}{\sqrt x-4}$ $= \frac{(\sqrt x-1)^2}{\sqrt x-2}-\frac{(4-\sqrt x)(4+\sqrt x)}{\sqrt x-4}$ $= \frac{(\sqrt x-1)^2}{\sqrt x-2}-\frac{-(\sqrt x-4)(\sqrt x+4)}{\sqrt x-4}$ $= \frac{(\sqrt x-1)^2}{\sqrt x-2}+(\sqrt x+4)$ $= \frac{(\sqrt x-1)^2+2(\sqrt x+4)(\sqrt x-2)}{2(\sqrt x-2)}$ $= \frac{x-2\sqrt x+1+2x-4\sqrt x+8\sqrt x-16}{2(\sqrt x-2)}$ $= \frac{3x+2\sqrt x-15}{2(\sqrt x-2)}$ $= \frac{3(\sqrt x-1)(\sqrt x+5)}{2(\sqrt x-2)}$