Toán 1) (x+2)(x-3) = 0 2) (x – 5)(7 – x) = 0 3) (2x +3)(-x + 7)=0 13/10/2021 By Emery 1) (x+2)(x-3) = 0 2) (x – 5)(7 – x) = 0 3) (2x +3)(-x + 7)=0
`1)` `(x+2)(x-3) = 0` `<=>` \(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\) Vậy phương trình trên có nghiệm `S={-2;3}` `2)` `(x – 5)(7 – x) = 0` `<=>` \(\left[ \begin{array}{l}x-5=0\\7-x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\) Vậy phương trình trên có nghiệm `S={5;7}` `3)` `(2x +3)(-x + 7)=0` `<=>` \(\left[ \begin{array}{l}2x+3=0\\-x+7\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x=-3\\-x=-7\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=7\end{array} \right.\) Vậy phương trình trên có nghiệm `S={-3/2;7}` Trả lời
$1)$ $(x+2)(x – 3) = 0$ \(\left[ \begin{array}{l}x+2 = 0\\x-3 =0\end{array} \right.\) \(\left[ \begin{array}{l}x= -2\\x= 3\end{array} \right.\) Vậy `x \in {-2;3}` $2)$ $(x – 5)(7 – x) = 0$ \(\left[ \begin{array}{l}x-5 =0\\7 – x =0\end{array} \right.\) \(\left[ \begin{array}{l}x= 5\\x = 7\end{array} \right.\) Vậy `x \in {5 ; 7}` $3)$ `(2x+3)(-x + 7) = 0` \(\left[ \begin{array}{l}2x +3 = 0\\-x + 7 =0\end{array} \right.\) \(\left[ \begin{array}{l}2x = -3\\-x = -7\end{array} \right.\) \(\left[ \begin{array}{l}x= \dfrac{-3}{2}\\x =7\end{array} \right.\) Vậy `x \in {-3/2 ; 7}` Trả lời
`1)` `(x+2)(x-3) = 0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
Vậy phương trình trên có nghiệm `S={-2;3}`
`2)` `(x – 5)(7 – x) = 0`
`<=>` \(\left[ \begin{array}{l}x-5=0\\7-x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
Vậy phương trình trên có nghiệm `S={5;7}`
`3)` `(2x +3)(-x + 7)=0`
`<=>` \(\left[ \begin{array}{l}2x+3=0\\-x+7\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x=-3\\-x=-7\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=7\end{array} \right.\)
Vậy phương trình trên có nghiệm `S={-3/2;7}`
$1)$
$(x+2)(x – 3) = 0$
\(\left[ \begin{array}{l}x+2 = 0\\x-3 =0\end{array} \right.\)
\(\left[ \begin{array}{l}x= -2\\x= 3\end{array} \right.\)
Vậy `x \in {-2;3}`
$2)$
$(x – 5)(7 – x) = 0$
\(\left[ \begin{array}{l}x-5 =0\\7 – x =0\end{array} \right.\)
\(\left[ \begin{array}{l}x= 5\\x = 7\end{array} \right.\)
Vậy `x \in {5 ; 7}`
$3)$
`(2x+3)(-x + 7) = 0`
\(\left[ \begin{array}{l}2x +3 = 0\\-x + 7 =0\end{array} \right.\)
\(\left[ \begin{array}{l}2x = -3\\-x = -7\end{array} \right.\)
\(\left[ \begin{array}{l}x= \dfrac{-3}{2}\\x =7\end{array} \right.\)
Vậy `x \in {-3/2 ; 7}`