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Suy ra -3=0 hoặc 4x+12=0
Mà -3 khác 0 suy ra 4x+12=0
suy ra 4x =-12
suy ra x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
(-3).(4x+12)=0
=>(-3).4+12.(-3)=0
tu day ta co:
=>-3.4x +(-36)=0
=>-3.4x=36
4x=36:(-3)
x=-12:4
=>x=-3
tick mk
=> (-3).4x+12. (-3)=0
-3. 4x+ (-36)=0
=>-3.4x=36
4x=36:(-3)
4x=-12
x=-12:4=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-4.\left(2x+9\right)-\left(-8x+3\right)-\left(x+13\right)=0\)
\(-8x-36+8x-3-x-13=0\)
\(-x-52=0\)
\(x=-52\)
k mk nha
thank you very much
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 100 - 7 ( x - 5 ) = 58
<=> 7 ( x - 5 ) = 100 - 58
<=> 7 ( x - 5 ) = 42
<=> x - 5 = 42 : 7
<=> x - 5 = 6
<=> x = 6 + 5
<=> x = 11
Tương tự tiếp.
a;100-7(x-5)=58
=>7(x-5)=100-58=42
=>x-5=42:7=6
=>x=6+5=11
b;12(x-1):3=72
=>12(x-1)=72.3=216
=>x-1=216:12=18
=>x=18+1=19
c;12-4(x-1)=4
=>4(x-1)=12-4=8
=>x-1=8:4=2
=>x=2+1=3
d;32-12x=8
=>12x=32-8=24
=>x=24:12=2
nho h do nhe viet moi tay lam day biet ko
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x + 8 chia hết cho x + 1
=> x + 1 + 7 chia hết cho x +1
=. x + 1 thuộc Ư ( 7 ) { - 1 , 1 , 7 , - 7 }
x + 1 = -1 =>x = -2
x + 1 = 1 => x = 0
x + 1 = 7 => x = 6
x + 1 = -7 => x = - 8
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy : \(\left(x-y^2+z\right)^2\ge0\forall x,y,z\)
\(\left(y-2\right)^2\ge0\forall y\)
\(\left(z+3\right)^2\ge0\forall z\)
Do đó : \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x-y^2+z=0\\y-2=0\\z+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2^2+\left(-3\right)=0\\y=2\\z=-3\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=7\\y=2\\z=-3\end{cases}}\)
Vậy : \(\left(x,y,z\right)=\left(7,2,-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\) - 92 = (443 - \(x\)) - 150
\(x-92\) = 443 - \(x\) - 150
\(x\) + \(x\) = 443 - 150 + 92
2\(x\) = 293 + 92
2\(x\) = 385
\(x\) = \(\dfrac{385}{2}\)
Vậy \(x=\dfrac{385}{2}\)
|48 - 3\(x\)| = 0 vì |48 - 3\(x\)| ≥ 0 ∀ \(x\) ⇒ |48 - 3\(x\)| = 0
⇔ 48 - 3\(x\) = 0
3\(x=48\)
\(x=48:3\)
\(x=16\)
Vậy \(x=16\)
=>3x-12=0 hoặc 8-x=0
=>x=4 hoặc x=8