(6x³ – 7x² – x + 2) : (2x + 1) – 3x (x – 2) 01/07/2021 Bởi Kylie (6x³ – 7x² – x + 2) : (2x + 1) – 3x (x – 2)
@Lạc $(6x³ – 7x² – x + 2) : (2x + 1) – 3x (x – 2)$ $= (6x³−6x²−x²+x−2x+2): (2x + 1) – 3x² + 6x $ $= [ 6x²(x-1) – x(x-1) – 2(x-1) ] : (2x + 1) – 3x² + 6x $ $= (x-1)(6x²-x-2) : (2x + 1) – 3x² + 6x $ $= (x-1)[3x.(2x+1)−2.(2x+1)] : (2x + 1) – 3x² + 6x $ $= (x-1)(2x+1)(3x-2) : (2x+1) – 3x² + 6x $ $ = (x-1)(3x-2) – 3x² + 6x $ $= 3x²−2x−3x+2−3x²+6x $ $ = x+2 $ Bình luận
$(6x^3-7x^2-x+2):(2x+1)-3x.(x-2)_{}$ $⇔\frac{6x^3-7x^2-x+2}{2x+1}-3x^2+6x_{}$ $⇔\frac{6x^3-6x^2-x^2+x-2x+2}{2x+1}-3x^2+6x_{}$ $⇔\frac{6x^2.(x-1)-x.(x-1)-2.(x-1)}{2x+1}-3x^2+6x_{}$ $⇔\frac{(x-1)(6x^2-x-2)}{2x+1}-3x^2+6x_{}$ $⇔\frac{(x-1)(6x^2+3x-4x-2)}{2x+1}-3x^2+6x_{}$ $⇔\frac{(x-1).[ 3x.(2x+1)-2.(2x+1)]}{2x+1}-3x^2+6x_{}$ $⇔\frac{(x-1)(2x+1)(3x-2)}{2x+1}-3x^2+6x_{}$ $⇔(x-1)(3x-2)-3x^2+6x_{}$ $⇔3x^2-2x-3x+2-3x^2+6x_{}$ $⇔x+2_{}$ Bình luận
@Lạc
$(6x³ – 7x² – x + 2) : (2x + 1) – 3x (x – 2)$
$= (6x³−6x²−x²+x−2x+2): (2x + 1) – 3x² + 6x $
$= [ 6x²(x-1) – x(x-1) – 2(x-1) ] : (2x + 1) – 3x² + 6x $
$= (x-1)(6x²-x-2) : (2x + 1) – 3x² + 6x $
$= (x-1)[3x.(2x+1)−2.(2x+1)] : (2x + 1) – 3x² + 6x $
$= (x-1)(2x+1)(3x-2) : (2x+1) – 3x² + 6x $
$ = (x-1)(3x-2) – 3x² + 6x $
$= 3x²−2x−3x+2−3x²+6x $
$ = x+2 $
$(6x^3-7x^2-x+2):(2x+1)-3x.(x-2)_{}$
$⇔\frac{6x^3-7x^2-x+2}{2x+1}-3x^2+6x_{}$
$⇔\frac{6x^3-6x^2-x^2+x-2x+2}{2x+1}-3x^2+6x_{}$
$⇔\frac{6x^2.(x-1)-x.(x-1)-2.(x-1)}{2x+1}-3x^2+6x_{}$
$⇔\frac{(x-1)(6x^2-x-2)}{2x+1}-3x^2+6x_{}$
$⇔\frac{(x-1)(6x^2+3x-4x-2)}{2x+1}-3x^2+6x_{}$
$⇔\frac{(x-1).[ 3x.(2x+1)-2.(2x+1)]}{2x+1}-3x^2+6x_{}$
$⇔\frac{(x-1)(2x+1)(3x-2)}{2x+1}-3x^2+6x_{}$
$⇔(x-1)(3x-2)-3x^2+6x_{}$
$⇔3x^2-2x-3x+2-3x^2+6x_{}$
$⇔x+2_{}$