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a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha

Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6

a) (2x+10)(x2+3)=0
(1) 2x+10=0 (2) x2+3=0
=> 2x=0+10 =>x2=0-3
=> 2x=10 =>x2=-3(Loại)
=>x=10:2
=>x=5
Vậy x=5
b)-12(x-5)+7(3-x)=5
-12.x+12.5+7.3+7.(-x)=5
-12.x+60+21+7.(-x)=5
-12.x+60+21+(-7).x=5
[(-12)+(-7)].x+60+21=5
-19x+81=5
-19x=5-81
-19x=-76
x=-76:(-19)
x=4
Vậy x=4.


1) \(10^{2n}-1=\left(10^n-1\right)\left(10^n+1\right)⋮12\left(v\text{ì}10^n-1⋮13\right)\)
bn giải cả bài giúp mk đc ko? đây là bài đội tuyển của mk!!!

a, ( - 168 ) + 72 . ( - 168 ) + ( - 168 ) . 27
= ( - 168 ) + ( - 12096 ) + ( - 4536 )
= - 12264 + - 4536
= - 16800
b, 22 . ( - 3 ) - ( 110 + 8 ) : ( - 3 )2
= 4 . ( - 3 ) - ( 1 + 8 ) : 9
= ( - 12 ) - 9 : 9
= ( - 12 ) - 1
= - 13
c, ( - 1075 ) - ( 29 - 1075 )
= ( - 1075 ) - ( - 1046 )
= - 29
d, ( - 9 ) + ( - 11 ) + 21 + ( - 1 )
= - 20 + 21 + ( - 1 )
= 1 + - 1
= 0
e, 30 + 12 + ( - 20 ) + ( - 12 ) - ( 30 - 20 ) + ( 12 - 12 )
= 42 + ( - 20 ) + ( - 12 ) - 10 + 0
= 22 + ( - 12 ) - 10 + 0
= 10 - 10 + 0
= 0 + 0
= 0
g, ( 13 - 135 + 49 ) - ( 13 + 49 )
= [( - 122 ) + 49 ] - 62
= ( - 73 ) - 62
= - 135
h, 35 - { 12 - [ ( - 14 ) + ( - 2 ) } ]
= 35 - { 12 - ( - 16 ) }
= 35 - 28
= 7
Bài 2:
a. x - 35 = ( - 12 ) - 3
x - 35 = - 15
x = - 15 + 35
x = 20
b, \(\frac{1}{4}\)+ \(\frac{1}{3}\): 3x = - 5
\(\frac{3}{12}+\frac{4}{12}\): 3x = - 5
\(\frac{7}{12}\): 3x = - 5
3x = \(\frac{7}{12}\): - 5
3x = \(\frac{-7}{60}\)
x = \(\frac{-7}{60}\): 3
x = \(\frac{-7}{180}\)
c,2x-1 = 8
2x-1 = 24
x = 4 + 1
x = 5
\(\left(13x-12^2\right):5=5\)
\(13x-12^2=25\)
\(13x-144=25\)
\(13x=169\)
\(x=13\)