Toán Giải hộ mình phương trình (2x+1)(3x-2)=(5x-8)(2x+1) 05/10/2021 By Harper Giải hộ mình phương trình (2x+1)(3x-2)=(5x-8)(2x+1)
$(2x+1)(3x-2)=(5x-8)(2x+1)$$⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0$$⇔(2x+1)[(3x-2)-(5x-8)]=0$$⇔(2x+1)(3x-2-5x+8)=0$$⇔(2x+1)(6-2x)=0$⇔\(\left[ \begin{array}{l}2x+1=0\\6-2x=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x=-1\\2x=6\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{-1}{2}\\x=3\end{array} \right.\) Trả lời
Đáp án + Giải thích các bước giải: `(2x+1)(3x-2)=(5x-8)(2x+1)` `<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0` `<=>(2x+1)(3x-2-5x+8)=0` `<=>(2x+1)(-2x+6)=0` `<=>` \(\left[ \begin{array}{l}2x+1=0\\-2x+6=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x=-1\\-2x=-6\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-\frac{1}{2}\\x=3\end{array} \right.\) Vậy `S={-(1)/(2);3}` Trả lời
$(2x+1)(3x-2)=(5x-8)(2x+1)$
$⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0$
$⇔(2x+1)[(3x-2)-(5x-8)]=0$
$⇔(2x+1)(3x-2-5x+8)=0$
$⇔(2x+1)(6-2x)=0$
⇔\(\left[ \begin{array}{l}2x+1=0\\6-2x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=-1\\2x=6\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-1}{2}\\x=3\end{array} \right.\)
Đáp án + Giải thích các bước giải:
`(2x+1)(3x-2)=(5x-8)(2x+1)`
`<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0`
`<=>(2x+1)(3x-2-5x+8)=0`
`<=>(2x+1)(-2x+6)=0`
`<=>` \(\left[ \begin{array}{l}2x+1=0\\-2x+6=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-1\\-2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\frac{1}{2}\\x=3\end{array} \right.\)
Vậy `S={-(1)/(2);3}`