Đáp án+Giải thích các bước giải:
`a)(x+4)^2+(3+x)(3+x)=1`
`⇔(x+4)^2+(3+x)^2=1`
`⇔x^2+8x+16+9+6x+x^2=1`
`⇔(x^2+x^2)+(8x+6x)+(16+9)=1`
`⇔2x^2+14x+25=1`
`⇔2x^2+14x+25-1=0`
`⇔2x^2+14x+24=0`
`⇔2(x^2+7x+12)=0`
`⇔x^2+7x+12=0`
`⇔(x^2+3x)+(4x+12)=0`
`⇔x(x+3)+4(x+3)=0`
`⇔(x+3)(x+4)=0`
\(⇔\left[ \begin{array}{l}x+3=0\\x+4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-3\\x=-4\end{array} \right.\)
Vậy `S=\{-3;-4\}`
`b)(x+4)/2-(x-1)/3=1`
`⇔(3(x+4))/6-(2(x-1))/6=6/6`
`⇔3(x+4)-2(x-1)=6`
`⇔3x+12-2x+2=6`
`⇔(3x-2x)+(12+2)=6`
`⇔x+14=6`
`⇔x=6-14`
`⇔x=-8`
Vậy `S=\{-8\}`
