1/ x.(x+ 7) = 0
2/ (x + 12).(x-3) = 0
3/ (- x + 5) . (3 – x) = 0
4/ (x- 1).(x + 2 ) . ( -3 ) = 0
1/ x.(x+ 7) = 0 2/ (x + 12).(x-3) = 0 3/ (- x + 5) . (3 – x) = 0 4/ (x- 1).(x + 2 ) . ( -3 ) = 0
By Eva
By Eva
1/ x.(x+ 7) = 0
2/ (x + 12).(x-3) = 0
3/ (- x + 5) . (3 – x) = 0
4/ (x- 1).(x + 2 ) . ( -3 ) = 0
Đáp án: ↓
Giải thích các bước giải:
1/ x(x+7) = 0
⇔┍ x = 0
┗ x+7 = 0
⇔┍ x = 0
┗ x = -7
vậy S ∈ {0; -7}
2/ (x+12).(x – 3) = 0
⇔ ┍ x+12=0
┗ x-3 = 0
⇔ ┍ x = -12
┗ x= 3
vậy S ∈ {-12;3}
3/ (- x+5).(3-x) = 0
⇔ ┍ -x+5=0
┗ 3-x = 0
⇔ ┍ -x =-5
┗ x = 3
vậy S ∈ {-5;3}
4/ (x-1).(x+2).(-3) = 0
⇔ ┍ x – 1= 0
┣ x+2 = 0
┗ -3 = -3
⇔ ┍ x = 1
┣ x= -2
┗ -3 = -3
vậy S ∈ {1;-2}
`a,x(x+7)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
`b,(x+12)(x-3)=0`
`⇔` \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
`c,(-x+5)(3-x)=0`
`⇔` \(\left[ \begin{array}{l}-x+5=0\\3-x=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
`d,(x-1)(x+2)(-3)=0`
`⇔(x-1)(x+2)=0`
⇔` \(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)