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

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`

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

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

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

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

a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)

\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)

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

\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)

\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)

\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))

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

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

\(=2\cdot\left(3-4\right)\)

\(=-2\)

b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)

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

\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)

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

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))

\(=3-1=2\)

c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)

\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

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

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))

\(=8-2\sqrt{15}\)

d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)

\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)

\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

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

\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))

\(=-3-\sqrt{6}\)

e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)

\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)

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

\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)

\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))

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

\(=\frac{16-12}{2}=\frac{4}{2}=2\)

f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))

\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)

\(=\sqrt{4+\sqrt{25}}\)

\(=\sqrt{4+5}=\sqrt{9}=3\)

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

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