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a)
\(|3x+1|=4\)
\(\Rightarrow\orbr{\begin{cases}3x+1=4\\3x+1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4-1\\3x=-4-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\div3\\x=-5\div3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1,6667\end{cases}}\)
Vậy x = 1
a, \(\left(x-1\right).\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b, \(\left(2x-4\right).\left(3x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\3x+9=0\end{matrix}\right.\left[{}\begin{matrix}2x=4\\3x=-9\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a) TH1: x-1=0 => x=1
TH2: x+2=0 => x=-2
b) TH1: 2x-4=0 <=> 2x= 4 <=> x=2
TH2: 3x+9=0 <=> 3x=-9 <=> x= -3
a, 7\(x\).(\(x\) - 10) = 0
\(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy \(x\in\) {0; 10}
b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0
\(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
a) \(\left(x^2+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\varnothing\\x=\pm2\end{cases}}}\)
Vậy x=\(\pm2\)
b) \(\left(x^3-27\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^3-27=0\\x^3+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^3=27\\x^3=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
Vậy x=3; x=-2
d) \(|3x+8|-|x-4|=0\)
\(\Leftrightarrow|3x+8|=|x-4|\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=x-4\\-3x-8=x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-12\\-4x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\x=-1\end{cases}}}\)
Vậy x=-6; x=-1
\(\frac{x-1}{2}=\frac{8}{x-1}\Leftrightarrow\left(x-1\right).\left(x-1\right)=2.8\Rightarrow\left(x-1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow x=4;-4\)
Câu b đang nghĩ
`\color{grey}\text{#071931}`
`(4 + 2x) * (9 - 3x) = 0`
TH1: `4 + 2x = 0 => 2x = -4 => x = -2`
TH2: `9 - 3x = 0 => 3x = 9 => x = 3`
Vậy, `x \in {-2; 3}`
____
`(3x - 19)^3 = 125`
`=> (3x - 19)^3 = 5^3`
`=> 3x - 19 = 5`
`=> 3x = 24`
`=> x = 8`
Vậy, `x = 8.`
a, - 2 .( x + 6 ) + 6 . ( x - 10 ) = 8
- 2x - 12 + 6x - 60 = 8
4x - 72 = 8
4x = 8 + 72
4x = 80
x = 20
b, - 4 . ( 2x + 9 ) - ( - 8x + 3 ) - ( x + 13 ) = 0
- 8x - 36 + 8x - 3 - x - 13 = 0
- x - 52 = 0
x = - 52
c, 7x . ( 2 + x ) - 7x . ( x + 3 ) = 14
7x . ( 2 + x - x - 3 ) = 14
7x . ( - 1 ) = 14
7x = 14 : ( - 1 )
7x = - 14
x = - 2
d, 2 . ( 5 + 3x ) + x = 31
10 + 6x + x = 31
10 + 7x = 31
7x = 31 - 10
7x = 21
x = 3
a)-2(x+6)+6(x-10)=8
-2x+-12+6x+-60=8
4x+-72=8
4x=80
x=80:4
x=20
b)-4(2x+9)-(-8x+3)-(x+13)=0
-8x+-36+8x-3+x-13=0
(-8x+8x)+-36+-3+x-13=0
0+-52+x=0
x=0-(-52)
x=-52
c)7x(2+x)-7x(x+3)=14
14x+7x2
Ta xét các TH sau để phá dấu GTTĐ :
TH 1 : Nếu \(x< -\frac{8}{3}\) thì ta có :
\(-\left(3x+8\right)-\left[-\left(x-4\right)\right]=0\)
\(\Leftrightarrow-2x=12\)
\(\Leftrightarrow x=-6\) ( thỏa mãn )
TH 2 : Nếu \(-\frac{8}{3}\le x< 4\) thì ta có :
\(\left(3x+8\right)-\left[-\left(x-4\right)\right]=0\)
\(\Leftrightarrow4x=-4\)
\(\Leftrightarrow x=-1\) ( thỏa mãn )
Th 3 : Nếu \(x\ge4\) thì ta có :
\(3x+8-\left(x-4\right)=0\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\) ( không thỏa mãn )
Vậy : \(x\in\left\{-1,-6\right\}\)
| 3x + 8 | - | x - 4 | = 0
<=> | 3x + 8 | = | x - 4|
Th1: 3x + 8 = x - 4
3x - x = -8 - 4
2x = - 12
x = -6
TH2: 3x + 8 = 4 - x
3x + x = 4 - 8
4x = - 4
x = - 1
Vậy x = -1 hoặc x = -6