Đáp án + Giải thích các bước giải:
`(5x-2)/(6)+(3-4x)/(2)=2-(x+7)/(3)`
`⇔(5x-2)/(6)+(3(3-4x))/(6)=(12)/(6)-(2(x+7))/(6)`
`⇒5x-2+3(3-4x)=12-2(x+7)`
`⇔5x-2+9-12x=12-2x-14`
`⇔5x-12x+2x=2-9+12-14`
`⇔-5x=-9`
`⇔x=(9)/(5)`
Vậy `S={(9)/(5)}`
Bài `4:`
`a//(x-5)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x-5=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\)
Vậy `S={5;-3}`
`b//3x(x-5)+6(x-5)=0`
`⇔(x-5)(3x+6)=0`
`⇔` \(\left[ \begin{array}{l}x-5=0\\3x+6=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
Vậy `S={5;-2}`
`c//2(x-3)-4(3-x)=0`
`⇔2(x-3)+4(x-3)=0`
`⇔(x-3)(2+4)=0`
`⇔6(x-3)=0`
`⇔x-3=0`
`⇔x=3`
Vậy `S={3}`
`d//x^{2}+5x+6=0`
`⇔(x^{2}+2x)+(3x+6)=0`
`⇔x(x+2)+3(x+2)=0`
`⇔(x+2)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\)
Vậy `S={-2;-3}`
