a: x(x-3)-2x+6=0
=>x(x-3)-2(x-3)=0
=>(x-3)(x-2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
b: \(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
=>\(15x^2-21x-25x+35+10x-15x^2+2-3x=4\)
=>-39x=4-35=-31
=>\(x=\dfrac{31}{39}\)
`a)x(x-3)-2x+6=0`
`=>x^2-3x-2x+6=0`
`=>x(x-3)-2(x-3)=0`
`=>(x-3)(x-2)=0=>[(x=3),(x=2):}`
`b)(3x-5)(5x-7)+(5x+1)(2-3x)=4`
`=>15x^2-21x-25x+35+10x-15x^2+2-3x=4`
`=>39x=33`
`=>x=11/13`
Ta có : x(x - 3) - 2x + 6 = 0
<=> x(x - 3) - (2x - 6) = 0
=> x(x - 3) - 2(x - 3) = 0
=> (x - 2)(x - 3) = 0
⇔\orbr{𝑥−2=0𝑥−3=0⇔\orbr{x−2=0x−3=0
⇔\orbr{𝑥=2𝑥=3⇔\orbr{x=2x=3
