\(\left(x-1\right)\left(x+2\right)\left(x-6\right)\left(x-3\right)=34\)
<=>\(\left[\left(x-1\right)\left(x-3\right)\right]\left[\left(x+2\right)\left(x-6\right)\right]-34=0\)
<=>\(\left(x^2-4x+3\right)\left(x^2+4x-12\right)-34=0\)
đặt x^2-4x+3=y ta đc:
\(y\left(y-15\right)-34=0\)
<=>\(y^2-15y-34=0\)<=>\(y^2-17y+2y-34=0\)
<=>\(y\left(y-17\right)+2\left(y-17\right)=0\)
<=>\(\left(y+2\right)\left(y-17\right)=0< =>\left[{}\begin{matrix}y+2=0\\y-17=0\end{matrix}\right.< =>\left[{}\begin{matrix}y=-2\left(1\right)\\y=17\left(2\right)\end{matrix}\right.\)
giải 1 : \(x^2-4x+3=-2< =>x^2-4x+5=0< =>x^2-4x+4+1=0< =>\left(x-2\right)^2=-1\left(vônghiem\right)\)
giải 2:\(x^2-4x+3=17< =>x^2-4x-14=0< =>x^2-4x+4-18=0< \left(x-2\right)^2=18< =>\left[{}\begin{matrix}x-2=-\sqrt{18}\\x-2=\sqrt{18}\end{matrix}\right.< =>\left[{}\begin{matrix}x=-\sqrt{18}+2\\x=\sqrt{18}+2\end{matrix}\right.\)
