a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

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

\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)

b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)

\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)

\(=-60-144\sqrt{2}+30\sqrt{2}+144\)

\(=84-114\sqrt{2}\)

a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(=\sqrt{3}-1-2-\sqrt{3}\)

=-3

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

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

=1

c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=3-\sqrt{6}+2\sqrt{6}-3\)

\(=\sqrt{6}\)

Giải:

\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)

\(=\left(\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\right).\left(\sqrt{6}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)

\(=\left(\dfrac{\sqrt{6}}{6}\right).\left(-\sqrt{2}\right).\left(-\sqrt{6}\right)\)

\(=\sqrt{2}\)

Vậy ...

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`

`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`

`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`

`=((9sqrt6)/6+(4sqrt6)/6)^2-24`

`=((13sqrt6)/6)^2-24`

`=13^2/6-24`

`=25/6`

ai giúp mk vs

\(VT=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right).\left(15+2\sqrt{6}\right)=201\)

     \(=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right).\left(15+2\sqrt{6}\right)\)

    \(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)\)

   \(=15^2-\left(2\sqrt{6}\right)^2=201=VP\)   (đpcm)

\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)\)

\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)\)

\(=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)\)

\(=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}\)

\(=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139\)

Ta có: \(-14,5\sqrt{2}\simeq-20,506\)

\(VT\ne VP\)

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