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đặt A=2^x +2^x+1 +.....+2^x+2021=2^x+2026-16
đặt 2A = 2^x+1 +2^x+2 +......+2^x+2022=2^x+2027-32
lấy 2A-A =2^x+2022-2^x=2^2026-16
vậy,ta suy ra x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(2^x\left(1+2+2^2+...+2^{2021}\right)=2^4\left(2^{2022}-1\right)\)
=>2^x=2^4
=>x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2^x+2^{x+1}+...+2^{x+2021}=2^{x+2026-16}\)
Đặt \(2A=2^{x+1}+2^{x+2}+...+2^{x+2022}=2^{x+2027+32}\)
Ta lấy \(2A-A=2^{x+2022}-2^x=2^{2026-16}\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(2VT=2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2022}\)
\(VT=2VT-VT=2^{x+2022}-2^x\)
\(\Rightarrow2^{x+2022}-2^x=2^{2026}-16\)
\(\Leftrightarrow2^{2022}.2^x-2^x=2^{2026}-2^4\)
\(\Leftrightarrow2^x\left(2^{2022}-1\right)=2^4\left(2^{2022}-1\right)\)
\(\Leftrightarrow2^x=2^4\Rightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(2^x-26=6\)
\(\Rightarrow2^x=6+26\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
2. \(64\cdot4^x=16^8\)
\(\Rightarrow4^3\cdot4^x=4^{16}\)
\(\Rightarrow4^x=4^{16}:4^3\)
\(\Rightarrow4^x=4^{13}\)
\(\Rightarrow x=13\)
3. \(\left(2x-1\right)^4=16\)
\(\Rightarrow\left(2x-1\right)^4=2^4\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
4. \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6x=-16+2x
=>6x-2x=-16
=>4x=-16
=>x=-4
Vậy x=4.
2(x-7)=-16
=>2x-14=-16
=>2x=-16+14
=>2x=-2
=>x=-1
Vậy x=-1.
2x : 2 =16
=> 2x =16 x 2
=>2x = 32
=> 2x = 25
=> x=5