yêu cầu của bài toán là j zậy bn

\(x^2+\left(16-x\sqrt{3}\right)^2=4\left(12-x\right)^2\)

\(\Leftrightarrow x^2+256-32\sqrt{3}x+3x^2=4\left(144-24x+x^2\right)\)

\(\Leftrightarrow4x^2-32\sqrt{3}x+256=576-96x+4x^2\)

\(\Leftrightarrow4x^2-4x^2-32\sqrt{3}x+96x+256-576=0\)

\(\Leftrightarrow\left(96-32\sqrt{3}\right)x-320=0\)

\(\Leftrightarrow\left(96-32\sqrt{3}\right)x=320\)

\(\Leftrightarrow x=\frac{320}{96-32\sqrt{3}}=\frac{15+5\sqrt{3}}{3}\)

2.5-X=1.3 suy ra X =1.2

2-5-X=-1.3 suy ra X=3.8

de vay ma ko bít

\(\Rightarrow2,5-x=1,3\)

\(\left|2,5-x\right|=1,3\)

\(\Leftrightarrow\orbr{\begin{cases}2,5-x=1,3\\2,5-x=-1,3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2,5-1,3\\x=2,5+1,3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)

  Vậy \(x\in\){1,2 ; 3,8}

\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)

\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)

\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)

\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)

\(=\left(-1\right)^{2018}+2018=2019\)

Để pt có 2 nghiệm dương:

\(\left\{{}\begin{matrix}\Delta'=\left(m-3\right)^2-\left(m-1\right)\ge0\\x_1+x_2=-2\left(m-3\right)>0\\x_1x_2=m-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-7m+10\ge0\\m< 3\\m>1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\ge5\\m\le2\end{matrix}\right.\\m< 3\\m>1\end{matrix}\right.\)

\(\Rightarrow1< m\le2\)

#)Giải :

\(\left(7x-11\right)^3=2^5.5^2+200\)

\(\left(7x-11\right)^3=32.25+200\)

\(\left(7x-11\right)^3=100=10^3\)

\(7x-11=10\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=3\)