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=>\(2^x\left(1+2+2^2+...+2^{2021}\right)=2^4\left(2^{2022}-1\right)\)
=>2^x=2^4
=>x=4
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
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 \(x=4\)
\(2VT=2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2022}\)
\(VT=2VT-VT=2^{x+2022}-2^x\)
\(\Rightarrow2^{x+2022}-2^x=2^{2026}-16\)
\(\Leftrightarrow2^{2022}.2^x-2^x=2^{2026}-2^4\)
\(\Leftrightarrow2^x\left(2^{2022}-1\right)=2^4\left(2^{2022}-1\right)\)
\(\Leftrightarrow2^x=2^4\Rightarrow x=4\)
a) x + 2006 = 2021
x= 2021 - 2006
x= 15
b) 2x - 2016 = 2 4 . 4
2x - 2016 = 64
2x = 64 + 2016
2x = 2080
x= 2080 : 2
x= 1040
c) 3. ( 2x + 1) ³ =81
( 2x-1)3 = 27
( 2x-1)3 = 33
=> 2x-1 = 3
2x= 2
x= 1
a, \(x\) + 2006 = 2021
\(x\) = 2021 - 2006
\(x\) = 15
3:
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(2x+1\right)^2+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)
\(\left(2x-3\right)^{2021}\left(2x-3-1\right)=0\)
\(\left(2x-3\right)^{2021}\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
Ta có:
\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)
\(\Rightarrow\left(2x-3\right)^{2022}-\left(2x-3\right)^{2021}=0\)
\(\Rightarrow\left(2x-3\right)^{2021}\left[\left(2x-3\right)-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x-3-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{3}{2};2\right\}\).
\(Toru\)
Ta có : ( 2x - 1 )2020 = ( 2x - 1 )2021
=> ( 2x - 1 )2021 - ( 2x - 1 )2020 = 0
=> ( 2x - 1 )2020 . [( 2x -1 )1 - 1 ] = 0
=> 2x - 1 = 0 2x = 1 x = 1/2
hoặc => =>
2x - 1 = 1 2x = 2 x =1
Vậy x = 1 hoặc x = 1/2