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a. 1440 : [ 120 - (3x + 9 ) ] = 120
120 - (3x + 9) = 1440 : 120 = 12
3x + 9 = 120 - 12 = 108
3x = 108 - 9 = 99
x = 99 : 3 = 33
b. 120 + [ ( 999 - 9x ) : 60 ] . 24 = 480
[ ( 999 - 9x ) : 60 ] . 24 = 480 - 120 = 360
( 999 - 9x ) : 60 = 360 : 24 = 15
( 999 - 9x ) = 15 . 60 = 900
9x = 999 - 900 = 99
x = 99 : 9 = 11
a, 2x -(-17) = 15
2x + 17 = 15
2x = 15 - 17
2x =(-2)
x = (-2) : 2
x = (-1)
a) x - (11 - x) = -48 + (-12 + x)
x - 11 + x = -48 + (-12) + x
x + x - x = -48 + (-12) + 11
x + 0 = (-60) + 11
x = -49
b) (15 - x) + (x - 12) = 7 - (-8 + x)
15 - x + x - 12 = 7 + 8 + x
15 - x + x - 12 = 15 + x
x + x - x = 15 - 15 + 12
x + 0 = 0 + 12
x = 12
c) (x - 12) - (2x + 31) = -6 - 5
x - 12 - 2x - 31 = -1
x - 12 - 2x = -1 + 31
x - 12 - 2x = 30
x - 2x = 30 + 12
-1x = 42
=> x = -42
d) |x + 5| - (-17) = 20
=> |x + 5| + 17 = 20
=> |x + 5| = 20 - 17
=> |x + 5| = 3
=> \(\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)
=> \(\orbr{\begin{cases}x=3-5\\x=-3-5\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)
12x2 - 3x = 0
=> 12x2 - 3x = (22 . 3x2) - 3x = 0
=> 3x . (4x - 1) = 0
=> 3x . 4x - 3x . 1 = 0
=> 3x . 4x - 3x = 0
=> 12x - 3x = 0
=> 4x =0
=> x = 0
a) Để(x^2-1).(2x-6)=0 thì 2x-6=0 suy ra x=3 và x^2-1=0 suy ra x=-1 hoặc 1
a) \(\left(x^2-1\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x^2-1=0\\2x-6=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x^2=1\\2x=6\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=3\end{array}\right.\)
Vậy \(x\in\left\{1;3\right\}\)
b) \(2x+3x-x-24=16\)
\(\Rightarrow2x+3x-x=16+24\)
\(\Rightarrow4x=40\)
\(\Rightarrow x=40:4=10\)
Vậy x = 10
c) \(\left(x^2+1\right)\left(x-5\right)\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x^2+1=0\\x-5=0\\x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x^2=-1\\x=0+5\\x=0+1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x\in\phi\\x=5\\x=1\end{array}\right.\)
Vậy \(x\in\left\{1;5\right\}\)
a) \(\left(x^2-1\right).\left(2x-6\right)=0\)
\(\Rightarrow\left(x^2-1\right).2\left(x-3\right)=0\)
\(\Rightarrow\left(x^2-1\right).\left(x-3\right)=0\)
\(\Rightarrow x^2-1=0\) hoặc \(x-3=0\)
+) \(x^2-1=0\Rightarrow x^2=1\Rightarrow x=1\) hoặc \(x=-1\)
+) \(x-3=0\Rightarrow x=3\)
Vậy \(x\in\left\{1;-1;3\right\}\)
b) \(2x+3x-x-24=14\)
\(\Rightarrow4x=40\)
\(\Rightarrow x=10\)
Vậy x = 10
c) \(\left(x^2+1\right).\left(x-5\right)\left(x-1\right)=0\)
\(\Rightarrow x^2+1=0\) hoặc \(x-5=0\) hoặc \(x-1=0\)
+) \(x^2+1=0\Rightarrow x^2=-1\) ( vô lí )
+) \(x-5=0\Rightarrow x=5\)
+) \(x-1=0\Rightarrow x=1\)
Vậy \(x\in\left\{5;1\right\}\)
a)(4-x)2+15=40
(4-x)2=40-15
(4-x)2=25
(4-x)2=52
=>4-x=5
x=5-4
x=1
a) ( 4 - x )2 + 15 = 40
( 4 - x )2 = 40 - 15
( 4 - x )2 = 25
( 4 - x )2 = 52
4 - x = 5
x = 4 - 5
x = ( - 1 )
b, ( 3x - 2 )10 = ( 3x - 2 )4
=> x \(\in\left\{0;1\right\}\)
c, X = 1 + 4 + 42 + ......... + 42017
4X = 4 + 41 + ......... + 42018
4X - X = ( 4 + 41 + ........ + 42018 ) - ( 1 + 4 + 42 + ......... + 42017 )
4X - X = 4 + 41 + ......... + 42018 - 1 - 4 - 42 - ......... - 42017
=> 3X = 42018 - 1
=> X = \(\frac{4^{2018}-1}{3}\)
\(a) (x-6).(-3x+15)=0\)
\(\Rightarrow x-6=0\) và \((-3x+15) =0\)
\(\Rightarrow x=6\) và \(-3x = -15\)
\(\Rightarrow x=6\) và \(x=5\)