GPT: 1) (3x-2)(4x+5)= 0 2) (2,3x-6,9)(0,1x+2)= 0 29/10/2021 Bởi Camila GPT: 1) (3x-2)(4x+5)= 0 2) (2,3x-6,9)(0,1x+2)= 0
`1) (3x-2)(4x+5)= 0` `⇔` \(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{array} \right.\) Vậy `S={2/3;-5/4}` `2)(2,3x-6,9)(0,1x+2)= 0` `⇔` \(\left[ \begin{array}{l}2,3x-6,9=0\\0,1x+2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=3\\x=20\end{array} \right.\) Vậy `S={3;20}` Bình luận
Đáp án: Giải thích các bước giải: 1) TH1: 3x-2=0 3x=2 x=$\frac{2}{3}$ TH2: 4x +5=0 4x=-5 x=$\frac{-5}{4}$ Vậy x∈ {$\frac{2}{3}$: $\frac{-5}{4}$} 2)TH1: 2,3x – 6,9=0 2,3x= 6,9 x=3 TH2: 0,1x +2=0 0,1x=-2 x= -20 Vậy x∈{3; -20} Bình luận
`1) (3x-2)(4x+5)= 0`
`⇔` \(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{array} \right.\)
Vậy `S={2/3;-5/4}`
`2)(2,3x-6,9)(0,1x+2)= 0`
`⇔` \(\left[ \begin{array}{l}2,3x-6,9=0\\0,1x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\\x=20\end{array} \right.\)
Vậy `S={3;20}`
Đáp án:
Giải thích các bước giải:
1) TH1:
3x-2=0
3x=2
x=$\frac{2}{3}$
TH2:
4x +5=0
4x=-5
x=$\frac{-5}{4}$
Vậy x∈ {$\frac{2}{3}$: $\frac{-5}{4}$}
2)TH1:
2,3x – 6,9=0
2,3x= 6,9
x=3
TH2:
0,1x +2=0
0,1x=-2
x= -20
Vậy x∈{3; -20}