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\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
a: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
b: \(=\sqrt{2}-1-\sqrt{2}=-1\)
\(1.\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}=\dfrac{\sqrt{3}+2+2-\sqrt{3}}{3-4}=-4\)\(2.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{8+6\sqrt{2}+6\sqrt{2}-8}{16-18}=\dfrac{-12\sqrt{2}}{2}-6\sqrt{2}\)\(3.\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.3+9}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\text{|}2\sqrt{2}-3\text{|}+\text{|}2\sqrt{2}+3\text{|}=4\sqrt{2}\)
\(4.\sqrt{29-4\sqrt{7}}-\sqrt{29+4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}.1+1}-\sqrt{28+2.2\sqrt{7}.1+1}=\sqrt{\left(2\sqrt{7}-1\right)^2}-\sqrt{\left(2\sqrt{7}+1\right)^2}=\text{|}2\sqrt{7}-1\text{|}-\text{|}2\sqrt{7}+1\text{|}=-2\)\(5.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7+2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\text{|}\sqrt{7}+1\text{|}-\text{|}\sqrt{7}-1\text{|}}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}\)
1)
\(\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}\)
\(=\dfrac{\left(\sqrt{3}+2\right)-\left(\sqrt{3}-2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}\)
\(=\dfrac{4}{\left(\sqrt{3}\right)^2-2^2}\)
\(=\dfrac{4}{3-4}=-4\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
a/ \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=7-4\sqrt{3}+7+4\sqrt{3}=14\)
a) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)
b) \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}+2}-\dfrac{12}{3-\sqrt{6}}=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}+2}\right)-\dfrac{12}{3-\sqrt{6}}\)
\(=\left(\dfrac{15\left(\sqrt{6}+2\right)+4\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}+2\right)}\right)-\dfrac{12}{3-\sqrt{6}}=\dfrac{15\sqrt{6}+30+4\sqrt{6}+4}{6+2\sqrt{6}+\sqrt{6}+2}-\dfrac{12}{3-\sqrt{6}}\) \(=\dfrac{34+19\sqrt{6}}{8+3\sqrt{6}}-\dfrac{12}{3-\sqrt{6}}=\dfrac{\left(34+19\sqrt{6}\right)\left(3-\sqrt{6}\right)-12\left(8+3\sqrt{6}\right)}{\left(8+3\sqrt{6}\right)\left(3-\sqrt{6}\right)}\)
\(=\dfrac{102-34\sqrt{6}+57\sqrt{6}-114-96-36\sqrt{6}}{24-8\sqrt{6}+9\sqrt{6}-18}=\dfrac{-108-13\sqrt{6}}{6+\sqrt{6}}\)
c) \(\sqrt{2+\sqrt{3}}+\sqrt{2+\sqrt{3}}=2\sqrt{2+\sqrt{3}}=\sqrt{2}.\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
câu này mk cảm thấy đề sai thì phải ; mà nếu o phải đề sai thì lời giải đó nha
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
\(a.\left(4+\sqrt{7}\right)\left(\sqrt{14}-\sqrt{2}\right)\sqrt{4-\sqrt{7}}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)\sqrt{7-2\sqrt{7}+1}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)^2=2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)=2\left(16-7\right)=18\) \(b.\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}=\dfrac{4\sqrt{2}+\sqrt{14}}{6+\sqrt{7+2\sqrt{7}+1}}+\dfrac{4\sqrt{2}-\sqrt{14}}{6-\sqrt{7-2\sqrt{7}+1}}=\dfrac{4\sqrt{2}+\sqrt{14}}{7+\sqrt{7}}+\dfrac{4\sqrt{2}-\sqrt{14}}{7-\sqrt{7}}=\dfrac{\left(4\sqrt{2}+\sqrt{14}\right)\left(7-\sqrt{7}\right)+\left(4\sqrt{2}-\sqrt{14}\right)\left(7+\sqrt{7}\right)}{49-7}=\dfrac{28\sqrt{2}-4\sqrt{14}+7\sqrt{14}-7\sqrt{2}+28\sqrt{2}+4\sqrt{14}-7\sqrt{14}-7\sqrt{2}}{42}=\dfrac{42\sqrt{2}}{42}=\sqrt{2}\)