Đáp án+Giải thích các bước giải:
`a)` `3(x-5)^2+2x(x-5)=0`
`=>3(x^2-10x+25)+2x^2-10x=0`
`=>3x^2-30x+75+2x^2-10x=0`
`=>(3x^2+2x^2)-(30x+10x)+75=0`
`=>5x^2-40x+75=0`
`=>5x^2-25x-15x+75=0`
`=>(5x^2-25x)-(15x-75)=0`
`=>5x(x-5)-15(x-5)=0`
`=>(x-5)(5x-15)=0`
`=>(x-5)(x-3).5=0`
`=>` $\left[\begin{matrix} x-5=0\\ x-3=0\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=5\\ x=3\end{matrix}\right.$
Vậy `x=5` hoặc `x=3`
`b)` `(x-3)^2-(x+3)(x-5)=3x+4`
`=>(x-3)^2-(x+3)(x-5)-3x-4=0`
`=>x^2-6x+9-x^2+2x+15-3x-4=0`
`=>(x^2-x^2)+(2x-6x-3x)+(9+15-4)=0`
`=>-7x+20=0`
`=>-7x=-20`
`=>` `x={-20}/{-7}={20}/{7}`
Vậy `x={20}/{7}`
`c)` `(2x-1)^2-(x+3)^2=0`
`=>4x^2-4x+1-x^2-6x-9=0`
`=>(4x^2-x^2)-(4x+6x)+(1-9)=0`
`=>3x^2-10x-8=0`
`=>3x^2-12x+2x-8=0`
`=>(3x^2-12x)+(2x-8)=0`
`=>3x(x-4)+2(x-4)=0`
`=>(x-4)(3x+2)=0`
`=>` $\left[\begin{matrix} x-4=0\\ 3x+2=0\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=4\\ 3x=-2\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=4\\ x=-2/3\end{matrix}\right.$
Vậy `x=4` hoặc `x=-2/3`
`d)` `4(2+3x)(3x-2)-(6x+1)^2=7`
`=>4(6x-4+9x-6x)-(36x^2-12x+1)-7=0`
`=>4(9x-4)-12x-1-7=0`
`=>36x^2-16-36x^2-12x-1-7=0`
`=>(36x^2-36x^2)-12x-(1+7+16)=0`
`=>-12x-24=0`
`=>-12x=24`
`=>x=24/{-12}=-2`
Vậy `x=-2`
