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30 tháng 7 2018

1. c)\(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)\)

\(\Leftrightarrow6+7\sqrt{6}-2\sqrt{6}-14\)

\(\Leftrightarrow-8+5\sqrt{6}\)

d)\(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)\)

\(\Leftrightarrow3-5\sqrt{3}+2\sqrt{3}-3\)

\(\Leftrightarrow-3\sqrt{3}\)

29 tháng 7 2018

@ngonhuminh @ngonhuminh

30 tháng 7 2018

a)   \(\left(\sqrt{12}-3\sqrt{75}\right).\sqrt{3}=\left(2\sqrt{3}-15\sqrt{3}\right).\sqrt{3}=-13\sqrt{3}.\sqrt{3}=-39\)

b)  \(\left(\sqrt{18}-4\sqrt{72}\right).2\sqrt{2}=\left(3\sqrt{2}-24\sqrt{2}\right).2\sqrt{2}=-21\sqrt{2}.2\sqrt{2}=-84\)

c)  \(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)=6+7\sqrt{6}-2\sqrt{6}-14=5\sqrt{6}-8\)

d)  \(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)=3-5\sqrt{3}+2\sqrt{3}-10=-7-3\sqrt{3}\)

31 tháng 7 2018

1.a) (\(\sqrt{12}\) -3\(\sqrt{75}\))\(\sqrt{3}\)

=\(\sqrt{12}\).\(\sqrt{3}\)-3\(\sqrt{75}\).\(\sqrt{3}\)

=\(2\sqrt{3}.\sqrt{3}-3.5\sqrt{3}.\sqrt{3}\)

=2.3-15.3

=6-45

= -39

b)\(\left(\sqrt{18}-4\sqrt{72}\right)2\sqrt{2}\)

\(\left(3\sqrt{2}-4.6\sqrt{2}\right).2\sqrt{2}\)

\(\left(3\sqrt{2}-24\sqrt{2}\right).2\sqrt{2}\)

\(3\sqrt{2}.2\sqrt{2}-24\sqrt{2}.2\sqrt{2}\)

= 6.2-48.2 = 12-96= -84

d)\(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)\)

\(3-5\sqrt{3}+2\sqrt{3}-10\)

\(-7-3\sqrt{3}\)

30 tháng 7 2018

\(\)c)\(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)\)

\(\Leftrightarrow6+6\sqrt{7}-2\sqrt{6}-14\)

\(\Leftrightarrow-8+5\sqrt{6}\)

d)\(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)\)

\(\Leftrightarrow3-5\sqrt{3}+2\sqrt{3}-3\)

\(\Leftrightarrow-3\sqrt{3}\)

26 tháng 6 2021

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)

\(=5-2\sqrt{2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

\(=2-2\sqrt{3}\)

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)

\(=-4\sqrt{5}+15\sqrt{2}\)

b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)

\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)

\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)

\(=8\sqrt{3}+2\sqrt{2}-4\)

c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)

\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)

=9-3

=6

d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

2 tháng 7 2021

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)