1 ) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(t=x^2+x\), ta được :
\(t^2+4t-12=0\)
\(\Leftrightarrow t^2-2t+6t-12=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
+ ) Khi \(t=2,\) thì :
\(x^2+x=2\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
+ ) Khi \(t=-6,\) thì :
\(x^2+x=-6\)
\(\Leftrightarrow x^2+x+6=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\) ( vô lí )
Vậy .........
2 ) \(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4-18x^3+13x^3-39x^2+x^2-3x-2x+6=0\)
\(\Leftrightarrow6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3+13x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(6x^2+x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(6x^2+3x-2x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left[3x\left(2x+1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(3x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)