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-12^2.x=56+10.13.x
<=>144x=56+130x
<=>144x-130x=56
<=>14x=56
<=>x=4
Vậy x=4
(-12)2.x=56+10.13.x
144x=56+130x
144x-130x=56
14x=56
X=4
Vay.......
`#3107.101107`
`1)`
`58 + 7x = 100`
`=> 7x = 100 - 58`
`=> 7x = 42`
`=> x = 42 \div 7`
`=> x = 6`
Vậy, `x = 6`
`2)`
`(x - 12) \div 12 = 12`
`=> x - 12 = 12*12`
`=> x - 12 = 144`
`=> x = 144 + 12`
`=> x = 156`
Vậy, `x = 156`
`3)`
`x - 56 \div 4 = 16`
`=> x - 14 = 16`
`=> x = 16 + 14`
`=> x = 30`
Vậy, `x = 30`
`4)`
`101 + (36 - 4x) = 105`
`=> 36 - 4x = 105 - 101`
`=> 36 - 4x = 4`
`=> 4x = 36 - 4`
`=> 4x = 32`
`=> x = 32 \div 4`
`=> x = 8`
Vậy, `x = 8`
`5)`
`2(x - 51) = 2*2^3 + 20`
`=> 2(x - 51) = 2^4 + 20`
`=> 2(x - 51) = 16 + 20`
`=> 2(x - 51) = 36`
`=> x - 51 = 36 \div 2`
`=> x - 51 = 18`
`=> x = 18 + 51`
`=> x = 69`
Vậy, `x = 69.`
1) 58 + 7x=100
7x = 100 - 58
7x = 42
x = 42 : 7
x = 6
2)(x-12):12=12
(x - 12) = 12 x 12
x - 12 = 144
x = 144 + 12
x = 156
3)x-56:4=16
x - 14 = 16
x = 16 + 14
x = 30
4) 101+(36-4x)=105
(36-4x) = 105 - 101
36-4x = 4
4x = 36 - 4
4x = 32
x = 32 : 4
x = 8
5)2(x-51)=2.2 mũ 3 + 20
2(x-51)=2.8 + 20
2(x-51)=16 + 20
2(x-51)= 36
x - 51 = 36 : 2
x - 51 = 18
x = 18 + 51
x = 69
\(\text{A)17×(x-8)=-54}\)
\(x-8=-54:17\)
\(x-8=\frac{-54}{17}\)
\(x=\frac{-54}{17}+8\)
\(x=\frac{82}{17}\)
\(\text{B)(12-2×x)÷4=-12}\)
\(12-2.x=-12:4\)
\(12-2.x=-3\)
\(2.x=12+3\)
\(2.x=15\)
\(\Rightarrow x=\frac{15}{2}\)
\(\text{C)9x+(-7)x=(-56)}\)
\(x.\left[9+\left(-7\right)\right]=-56\)
\(x.2=-56\)
\(x=-56:2\)
\(x=-28\)
\(\text{D)(3x -9)×(x+12)=0}\)
\(\Rightarrow\orbr{\begin{cases}3x-9=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=9\\x=-12\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-12\end{cases}}}\)
\(\Rightarrow x\in\left\{3;-12\right\}\)
\(\text{E) (x+4)^3=-8}\)
\(\left(x+4\right)^3=2^3\)
\(\Rightarrow x+4=2\)
\(x=2-4\)
\(x=-2\)
\(\text{F)(x)-7=11}\)
mk ko hiểu đề bài
Th1 , \(x-7=11\)
\(x=11+7\)
\(x=18\)
TH2, \(|x|-7=11\)
\(|x|=11+7\)
\(|x|=18\)
\(\Rightarrow x\in\left\{\pm18\right\}\)
học tốt
E) (x + 4)3 = -8
=> (x+4)3 = (-2)3
=> x + 4 = -2
=> x = -2 - 4
=> x = -6
Vậy x = -6
F) Cái chỗ (x) có phải là giá trị tuyệt đối không ạ ? Mình sửa đề bài như sau :
|x| - 7 = 11
\(\Rightarrow\orbr{\begin{cases}x-7=11\\x-7=-11\end{cases}\Rightarrow\orbr{\begin{cases}x=11+7\\-11+7\end{cases}\Rightarrow}\orbr{\begin{cases}x=18\\x=-4\end{cases}}}\)
Vậy x = 18 hoặc x= -4
a 25 + 8 + 12 + 25 = 70 b 87 + [ - 750 ] + 2018 - 13 + 750 = 2842 c [ 45 + 12 - 13 ] - [ 45 - 13 + 12 ] = 90 minh tinh gom bn ah
a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)
\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)
\(=a^{12}\cdot b^3\)
b) \(b^6\cdot b\cdot c^7\cdot c^8\)
\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)
\(=b^7\cdot c^{15}\)
c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)
\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)
\(=a^{18}\cdot c^{21}\)
d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)
\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)
\(=a^5\cdot b^4\cdot c^4\)
a) Kiểm tra lại nhé
b) \(b^6.b^7.c^8\)
\(=b^{6+7}.c^8=b^{13}.c^8\)
c) \(a^8.a^9.a.c.c^{20}\)
\(=a^{8+9+1}.c^{1+20}\)
\(=a^{18}.c^{21}\)
d) \(a^2.a^3.b^4.c.c^3\)
\(=a^{2+3}.b^4.c^{1+3}\)
\(=a^5.b^4.c^4\)
\(#WendyDang\)
= 144 . x = 56 + 130. x
= 144 . x - 130 . x = 56
= x.[144 - 130] = 56
= x.14 = 56
= x = 56 : 14
= x = 4