`a)`
`(x-3).(x+2)+(x-1).(x+5)=11`
`⇔x^2-x-6+x^2+4x-5=11`
`⇔2x^2+3x-22=0`
`⇔(x-(\sqrt(185)-3)/4)(x-(-\sqrt(185)-3)/4)=0`
`⇔`\(\left[ \begin{array}{l}x=(\sqrt(185)-3)/4\\x=(-\sqrt(185)-3)/4)\end{array} \right.\)
`b)`chưa sửa đề
`(3x-2).(x+1)-(x-2)-(3x+5)=32`
`⇔3x^2+x-2-x+2-3x-5-32=0`
`⇔3x^2-3x-37=0`
`⇔(x-(\sqrt(453)+3)/6)(x-(\sqrt(453)-3)/6)=0`
`⇔`\(\left[ \begin{array}{l}x=(\sqrt(453)+3)/6\\x=(\sqrt(453)-3)/6)\end{array} \right.\)
`b)`đã sửa đề :
`(3x-2).(x+1)-(x-2).(3x+5)=32`
`⇔3x^2+x-2-3x^2+x+10-32=0`
`⇔2x-24=0`
`⇔2(x-12)=0`
`⇔x-12=0`
`⇔x=12`
