( x – 2 )($x^{2}$ + 3x -2) = $x^{3}$ – 8 08/09/2021 Bởi Remi ( x – 2 )($x^{2}$ + 3x -2) = $x^{3}$ – 8
`(x-2)(x^2+3x-2)=x^3-8``⇔(x-2)(x^2+3x-2)=(x-2)(x^2+2x+4)``⇔(x-2)(x^2+3x-2)-(x-2)(x^2+2x+4)=0``⇔(x-2)(x^2+3x-2-x^2-2x-4)=0``⇔(x-2)(x-6)=0` ⇔\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\) `Vậy S={2;6}` Bình luận
Đáp án: `S={2;6}` Giải thích các bước giải: `(x-2)(x²+3x-2)=x³-8` `<=> (x-2)(x²+3x-2)-(x³-8)=0` `<=> (x-2)(x²+3x-2)-(x-2)(x²+2x+4)=0` `<=> (x-2)(x²+3x-2-(x²+2x+4))=0` `<=> (x-2)(x²+3x-2-x²-2x-4)=0` `<=> (x-2)(x-6)=0` `<=>`\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\) Vậy `S={2;6}` Bình luận
`(x-2)(x^2+3x-2)=x^3-8`
`⇔(x-2)(x^2+3x-2)=(x-2)(x^2+2x+4)`
`⇔(x-2)(x^2+3x-2)-(x-2)(x^2+2x+4)=0`
`⇔(x-2)(x^2+3x-2-x^2-2x-4)=0`
`⇔(x-2)(x-6)=0`
⇔\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\)
`Vậy S={2;6}`
Đáp án: `S={2;6}`
Giải thích các bước giải:
`(x-2)(x²+3x-2)=x³-8`
`<=> (x-2)(x²+3x-2)-(x³-8)=0`
`<=> (x-2)(x²+3x-2)-(x-2)(x²+2x+4)=0`
`<=> (x-2)(x²+3x-2-(x²+2x+4))=0`
`<=> (x-2)(x²+3x-2-x²-2x-4)=0`
`<=> (x-2)(x-6)=0`
`<=>`\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\)
Vậy `S={2;6}`