Đáp án:
Giải thích các bước giải:
`a) 6x^2 -(2x-3)(3x+2)=1`
`6x^2 -(2x.3x +2x.2 -3.3x-3.2)=1`
`6x^2 -(6x^2 +4x -9x -6)=1`
`6x^2 - 6x^2 +5x+6=1`
`(6x^2 -6x^2) +5x+6=1`
`5x +6=1`
`5x =1-6`
`5x =-5`
`x=-1`
`b)(x+1)^3 -(x-1)(x^2 +x+1)-2=0`
`x^3 +3x^2 +3x + 1^3 - (x^3 -1^3) -2=0`
`x^3 +3x^2 +3x +1 -x^3 +1 -2=0`
`(x^3 -x^3) +(1+1-2) +3x^2 +3x=0`
`3x^2 +3x=0`
`3x(x+1)=0`
$\left[\begin{matrix} 3x=0\\ x+1=0\end{matrix}\right.$
$\left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.$
`c)x^3 -3x^2 +3x-1=0`
`(x-1)^3 =0`
`x-1=0`
`x=1`
`d) 4y^2 +12y +25 +8x+x^2 =0`
`4y^2 + 12y + 9 + 16 +8x +x^2 =0`
`(4y^2 +12y +9 )+(16+8x+x^2)=0`
`[(2y)^2 +2.2y.3 +3^2]+[4^2 +2.4.x +x^2]=0`
`(2y+3)^2 + (4+x)^2=0`
Ta thấy `(2y+3)^2 ≥ 0 ∀ y`
Dấu ` ''=''` xảy ra `<=>(2y+3)^2 =0`
`2y +3=0`
`y= -3/2`
`(4+x)^2 ≥0 `
Dấu ` ''='' ` xảy ra `<=>(4+x)^2 =0`
`4+x =0`
`x=-4`
Vậy `x = -3/2 , y=-4`
