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1)
a)-24+3(x-4)=111
3(x-4)=111-(-24)
3(x-4)=111+24
3(x-4)=135
x-4=135:3
x-4=45
x =45+4
x =49
b)(2x-4)(3x+63)=0
\(\Rightarrow\)\(\orbr{\begin{cases}2x-4=0\\3x+63=0\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=2\\x=-21\end{cases}}\)
Vậy x\(\in\){2;-21}
c)|x-7|-4=(-2)4
|x-7| =(-2)4+4
|x-7| =16+4
|x-7| =20
\(\Rightarrow\)\(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=14\\x=0\end{cases}}\)
Vậy x\(\in\){14;0}
d)(x-1)2=144
(x-1)2=122
\(\Rightarrow\)x-1=12
x =12+1
x =13
e)(x+7)3=-8
(x+7)3=(-2)3
\(\Rightarrow\)x+7=-2
x =-2-7
x =-9
2)
a)Ta có:
\(3n+12⋮n-3\)
\(\Rightarrow3n-9+21⋮n-3\)
\(\Rightarrow3\left(n-3\right)+21⋮n-3\)
\(\Rightarrow21⋮n-3\)
\(\Rightarrow n-3\inƯ\left(21\right)\)
\(\Rightarrow n-3\in\left\{1;3;7;21\right\}\)
Ta có bảng sau:
n-3 | 1 | 3 | 7 | 21 |
n | 4 | 6 | 10 | 24 |
Vậy\(n\in\left\{4;6;10;24\right\}\)
b)Ta có:
\(n+9⋮n-1\)
\(\Rightarrow n-1+10⋮n-1\)
\(\Rightarrow10⋮n-1\)
\(\Rightarrow\)\(n-1\inƯ\left(10\right)\)
\(\Rightarrow n-1\in\left\{1;2;5;10\right\}\)
Ta có bảng sau:
n-1 | 1 | 2 | 5 | 10 |
n | 2 | 3 | 6 | 11 |
Vậy \(n\in\left\{2;3;6;11\right\}\)
107.10x+3=217.517
107.10x+3=(2.5)17
107.10x+3=1017
107.10x+3=107.1010
=>10x+3=1010(vì107=107)
=>x+3=10
x =10-3=7
vậy x=7
x2017=x
=>x=1
vì 1 mũ 10,124,24,45,347,87,...thì luôn luôn bằng chính nó (bằng 1)
a) 3x=9
Ta có: 9= 32
Mà 3x =32
=> x=2
b)x3=27
Ta có: 27 = 3.3.3= 33
Mà x3 =33
=> x=3
c)3.2x=24
=> 2x= 24: 3
=> 2x= 8
Mà 8= 23 => 2x= 23
Vậy x= 3
b) \(11\cdot6^{x-1}+2\cdot6^{x+1}=11\cdot6^{11}+2\cdot6^{13}\)
\(\Leftrightarrow11\cdot6^{x-1}+2\cdot6^{x+1}=11\cdot6^{12-1}+2\cdot6^{12+1}\)
\(\Leftrightarrow x=12\)
A)3.5n.52+4.5n:53=19.9765625
5n(3.52+4:53)=185546875
5n.\(\frac{12}{5}\)=185546875
b)11.6x-1+2.6x+1=11.611+2.613
11.6x-1+2.6x+1 = 11. 612-1+ 2. 612+1
=> x= 12
c) 24-x / 165 = 326
24-x / 220= 230
24-x = 250
=> 4-x = 50
x= -46
c) \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(2^{4-x}:2^{20}=\left(2^5\right)^6\)
\(2^{4-x}=2^{30}.2^{20}\)
\(2^{4-x}=2^{50}\)
=> \(4-x=50\)
=> \(x=4-50=-46\)
vậy x = -46
\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
<=> \(\left(19x+50\right):14=25-16\)
<=> \(\left(19x+50\right):14=9\)
<=> \(19x+50=126\)
<=> \(19x=76\)
<=> \(x=4\)
câu B làm j có \(x\)để tìm
\(390-\left(x-7\right)=169:13\)
<=> \(390-x+7=13\)
<=> \(390-x=6\)
<=> \(x=384\)
\(x-6:2-\left(48:24\right):2:6-3=0\)
<=> \(x-3-2:2:6=3\)
<=> \(x-3-\frac{1}{6}=3\)
<=> \(x=\frac{37}{6}\)
\(x+5.2-\left(32+16.3:6-15\right)=0\)
<=> \(x+10-25=0\)
<=> \(x=15\)
\(2^x+2^{x+1}=24\)
\(2^x+2^x\cdot2=24\)
\(2^x\cdot\left(1+2\right)=24\)
\(2^x\cdot3=24\)
\(2^x=8=2^3\)
Vậy x = 3
\(2^x+2^{x+1}=24\Rightarrow2^x\left(1+2\right)=24\Rightarrow2^x=8\Rightarrow x=3\)