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a: \(3^x-2=2^7\)
\(\Leftrightarrow3^x=128+2=130\)(vô lý)
b: \(4^{x+1}=64\)
=>x+1=3
hay x=2
c: \(\left(5x+1\right)^2=1^{2016}=1\)
=>5x+1=1 hoặc 5x+1=-1
=>x=0 hoặc x=-2/5
d: \(2^{2\left(x-1\right)}=8\)
=>2(x-1)=3
=>x-1=3/2
hay x=5/2

12 + ( 5 + x ) = 20 5.22 + ( x + 3 ) = 52 23 + ( x + 3 ) = 52 43 - ( x - 2 ) = 52
17 + x = 20 5.4 + x + 3 = 25 8 + x + 3 = 25 64 - x + 2 = 25
x = 20 - 17 20 + 3 + x = 25 11 + x = 25 66 - x = 25
x = 3 23 + x = 25 x = 25 - 11 x = 66 - 25
x = 25 - 23 x = 14 x = 41
x = 2
Đăng nhìu v bn :) Đáng quan ngại đây :)


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\(3\cdot7^{x+1}+7^{x+2}=490\\ \Rightarrow3\cdot7^x\cdot7+7^x\cdot7^2=490\\ \Rightarrow21\cdot7^x+7^x\cdot49=490\\ \Rightarrow7^x\cdot\left(21+49\right)=490\\ \Rightarrow7^x\cdot70=490\\ \Rightarrow7^x=490:70\\ \Rightarrow7^x=7\\ \Rightarrow x=1\)
Vậy: \(x=1\)
`3*7^(x+1)+7^(x+2)=490`
`=>3*7^x*7+7^x*7^2=490`
`=>21*7^x+7^x*49=490`
`=>7^x*(21+49)=490`
`=>7^x*70=490`
`=>7^x=490:70`
`=>7^x=7`
`=>7^x=7^1`
`=>x=1`