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để (2x - 6) x (3x - 12) = 0
thì (2x - 6) = 0 và (3x - 12) = 0
2x - 6 = 0
=> 2x = 0 = 6 = 6 => x = 6 : 2 = 3
3x - 12 = 0
=> 3x = 0 + 12 = 12 => x = 12 : 3 = 4
Vậy x = 3 và x = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a, => x-12 = 0
=> x=12
b, => 27-x =1
=> x= 27-1 -26
c, => 2x = 69(4-2) =69.2
=> x=69
d, => x-12=0
=> x=12
a) (x - 12) . 105 = 0
=> x - 12 = 0
=> x = 12
b) 47 . (27 - x) = 47
=> 27 - x = 1
=> x = 27 - 1 = 26
c) 2x + 69 . 2 = 69 . 4
=> 2x = 69 . 4 - 69 . 2
=> 2x = 69 . (4 - 2) = 69 . 2
=> x = 69 . 2 : 2
=> x = 69
d) 2x - 12 - x = 0
=> x - 12 = 0
=> x = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
2x - 12 - x = 0
2x - x - 12 = 0
x - 12 = 0
x = 0 + 12
x = 12
Vậy ....
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a) \(6x+15\times8=12\times\left(19-x\right)\)
\(6x+120=228-12x\)
\(6x+120-228+12x=0\)
\(18x-108=0\)
\(18x=108\)
\(x=6\)
b) \(160-\left(35\div x+3\right)\times15=15\)
\(160-\left(35\div x+3\right)=1\)
\(35\div x+3=159\)
\(35\div x=156\)
\(x=\dfrac{35}{156}\)
c) \(2x-\left(1309\div11-19\right)-2=0\)
\(2x-1309\div11-19=2\)
\(2x-119-19=2\)
\(2x-119=21\)
\(2x=140\)
\(x=70\)
d) \(\left(x-7\right)\times\left(2x-16\right)=0\)
\(x-7=0;2x-16=0\)
\(x=7;2x=16\)
\(x=7;x=8\)
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\(a,\left(4x-8\right)\left(12-y\right)=0\)
\(\Rightarrow\hept{\begin{cases}4x-8=0\\12-y=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}4x=8\\y=12\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=12\end{cases}}\)
\(b,\left(2x-12\right)-14=86\)
\(\Rightarrow2\left(x-6\right)=86+14=100\)
\(\Rightarrow x-6=50\)
\(\Rightarrow x=50+6=56\)
\(c,\left(127-x\right)-17=43\)
\(\Rightarrow127-x=43+17=60\)
\(\Rightarrow x=127-60=67\)
\(\left(4x-8\right)\left(12-y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-8=0\\12-y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}4x=8\\y=12-0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\y=12\end{cases}}}\)
Vậy x = 2 hoặc y = 12
\(\left(2x-12\right)-14=86\)
\(\Leftrightarrow2x-12=100\)
\(\Leftrightarrow2x=112\)
\(\Leftrightarrow x=56\)
Vậy x = 56
\(\left(127-x\right)-17=43\)
\(\Leftrightarrow127-x=60\)
\(\Leftrightarrow x=127-60\)
\(\Leftrightarrow x=67\)
Vậy x = 67
![](https://rs.olm.vn/images/avt/0.png?1311)
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2x + ( 5 - x) + 3* ( x - 79 ) = 0
2x + 5 - x + 3x - 3*79 = 0
2x - x + 3x + 5 - 237 = 0
4x - 232 = 0
4x = 232
x = 232 : 4
x = 58
~ Thiên mã ~
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a) \(x^3-16x=0\)
⇔\(x\left(x^2-16\right)=0\)
⇒\(x=0\) hoặc \(x^2-16=0\)
\(TH_1:x=0\)
\(TH_2:x^2-16=0\) ⇔ \(x^2=16\) ⇔ \(x=\pm4\)
Vậy \(x\in\left\{0;\pm4\right\}\)
b) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
⇒ \(2x+1=x-1\)
⇒ \(2x+2=x\)
⇒ \(2\left(x+1\right)=x\) ⇒ x = -2
Vậy x = -2
Đáp án: x = 3
tìm x, biết (2x-8) (2x+12)=0
ta có
TH1: 2x-8=0=> 2x=8=>2x=24=> x =4
TH2: 2x-8=0=> 2x=8=>2x=24=> x =-4
TH3: 2x+ 12=0=> 2x= -12 => x= -6
vậy x thuộc { 4;-4;-6}