a) x(x-4)+1=3x-5
=>\(x^2-4x+1-3x+5=0\)
=>\(x^2-7x+6=0\)
=>\(x^2-5x-2x+1+5=0\)
=>\(x^2-2x+1-5x+5=0\)
=>\(\left(x-1\right)^2-5\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-1-5\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-1-5=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)
b)
\(2x^3-3x^2-2x+3=0\)
=>\(2x\left(x^2-1\right)-3\left(x^2-1\right)=0\)
=>\(\left(x^2-1\right)\left(2x-3\right)=0\)
=>\(\left[{}\begin{matrix}x^2-1=0\\x^2-1=0\\2x-3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{3}{2}\end{matrix}\right.\)
a:=>x^2-4x+1=3x-5
=>x^2-7x+6=0
=>(x-1)(x-6)=0
=>x=1 hoặc x=6
b: =>x^2(2x-3)-(2x-3)=0
=>(2x-3)(x-1)(x+1)=0
hay \(x\in\left\{\dfrac{3}{2};1;-1\right\}\)
x2 ( 2x3 – 4x + 3)
