Đáp án+Giải thích các bước giải:
`a) 3x(x-2)-x+2=0`
`<=>3x(x-2)-(x-2)=0`
`<=>(x-2)(3x-1)=0`
`<=>`$\left[\begin{matrix} x-2=0\\ 3x-1=0\end{matrix}\right.$
`<=>`$\left[\begin{matrix} x=2\\ x=\dfrac{1}{3}\end{matrix}\right.$
Vậy `x=2` hoặc `x=1/3`
`b) x(x-4)-2x+8=0`
`<=>x(x-4)-2(x-4)=0`
`<=>(x-4)(x-2)=0`
`<=>`$\left[\begin{matrix} x-4=0\\ x-2=0\end{matrix}\right.$
`<=>`$\left[\begin{matrix} x=4\\ x=2\end{matrix}\right.$
Vậy `x=4` hoặc `x=2`
`c) x^2-25-(x+5)=0`
`<=>(x-5)(x+5)-(x+5)=0`
`<=>(x+5)(x-5-1)=0`
`<=>(x+5)(x-6)=0`
`<=>`$\left[\begin{matrix} x+5=0\\ x-6=0\end{matrix}\right.$
`<=>`$\left[\begin{matrix} x=-5\\ x=6\end{matrix}\right.$
Vậy `=-5` hoặc `x=6`
`d) (2x-1)^2-(4x^2-1)=0`
`<=>(2x-1)^2-(2x-1)(2x+1)=0`
`<=>(2x-1)(2x-1-2x-1)=0`
`<=>(2x-1).(-2)=0`
`<=>2x-1=0`
`<=>x=1/2`
Vậy `=1/2`
`e) (3x-1)^2-(x+5)^2=0`
`<=>(3x-1-x-5)(3x-1+x+5)=0`
`<=>(2x-6)(4x+4)=0`
`<=>2.(x-3).4(x+1)=0`
`<=>(x-3)(x+1)=0`
`<=>`$\left[\begin{matrix} x-3=0\\ x+1=0\end{matrix}\right.$
`<=>`$\left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.$
Vậy `x=3` hoặc `x=-1`
`f) x^3-8-(x-2)(x-12)=0`
`<=>(x^3-8)-(x-2)(x-12)=0`
`<=>(x-2)(x^2+2x+4)-(x-2)(x-12)=0`
`<=>(x-2)(x^2+2x+4-x+12)=0`
`<=>(x-12)(x^2+x+16)=0`
Ta có: `x^2+x+16`
`=(x+1/2)^2+63/4`
Với mọi `x` có: `(x+1/2)^2+63/4>=63/4>0`
hay `x^2+x+16>0`
Suy ra `x-12=0`
`<=>x=12`
Vậy `x=12`
