Tìm x biết:
a)-12+(-9+x)=0
b)21+(25-x)=2
c)461+(x-45)=387
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x - 36 : 18 = 12
x - 2 = 12
x= 14
(x - 36) : 18 = 12
(x - 36) = 216
x = 216 + 36
x = 352
461 + (x - 45) = 387
461 + x - 45 = 387
x = 387 - 461 + 45
x = -29
(x-36):18=12
x-36=12x18
x-36=216
x=216+36
x= 252
x-36:18=12
x-2=12
x=14
461+(x-45)=387
x-45= 387 - 461
x-45=-74
x=-74+45
x=-29
461 + (x - 45) = 387
x - 45 = 387 - 461
x - 45 = -74
x = -74 + 45
x = -29
11 - (-53 + x) = 97
-53 + x = 11 - 97
-53 + x = -86
x = -86 - -53
x = -33
- (x+84) + 213 = -16
x - 84 + 213 = -16
x - 84 = -16 - 213
x - 84 = -229
x = -229 + 84
x = -145
461 + ( x - 45 ) = 387
x - 45 = 387 - 461
x - 45 = - 74
x = - 74 + 45
x = - 29
Vậy x = - 29
a: =>7x=42
hay x=6
b: =>5x=35
hay x=7
c: =>x-14=16
hay x=30
d: =>36-4x=4
=>4x=32
hay x=8
e: =>x-12=144
hay x=156
f: =>3x-16=14
hay x=10
g: =>x+33=45
hay x=12
h: =>(x+9):2=39
=>x+9=78
hay x=69
a: =>7x=42
hay x=6
b: =>5x=35
hay x=7
c: =>x-14=16
hay x=30
d: =>36-4x=4
=>4x=32
a, \(x+45=54\Leftrightarrow x=9\)
b, \(x+9=3\Leftrightarrow x=-6\)
c, \(5^{2x}=5^2\Rightarrow2x=2\Leftrightarrow x=1\)
\(262+\left(2x-123\right)=283\\ \Rightarrow2x-123=21\\ \Rightarrow2x=144\\ \Rightarrow x=72\)
\(213-\left(x+84\right)=-16\\ \Rightarrow-\left(x+84\right)=-229\\ \Rightarrow x+84=229\\ \Rightarrow x=145\)
\(416+\left(x-45\right)=387\\ \Rightarrow x-45=-29\\ \Rightarrow x=16\)
\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)
\(b,\left(10-x\right)\left(3-x\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
a) -12 + ( -9 + x ) = 0
( -9 + x ) = 0 + 12
( -9 + x ) = 12
x = 12 + 9
x = 21
Vậy x = 21
b) 21 + ( 25 - x ) = 2
( 25 - x ) = 2 - 21
( 25 - x ) = -19
x = 25 + 19
x = 44
Vậy x = 44
c) 461 + ( x - 45 ) = 387
( x - 45 ) = 387 - 461
( x - 45 ) = -74
x = - 74 + 45
x = -29
Vậy x = -29
a, => -12-9+x = 0
=> x-21 = 0
=> x=0+21 = 21
b, => 21+25-x = 2
=> 46-x = 2
=> x = 46-2 = 44
c, => 461+x-45 = 387
=> 416+x = 387
=> x = 387 - 416 = -29
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