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

=>\(2^x\left(1+2+2^2+...+2^{2021}\right)=2^4\left(2^{2022}-1\right)\)

=>2^x=2^4

=>x=4

Đặt \(A=2^x+2^{x+1}+...+2^{x+2021}=2^{x+2026-16}\)

Đặt \(2A=2^{x+1}+2^{x+2}+...+2^{x+2022}=2^{x+2027+32}\)

Ta lấy \(2A-A=2^{x+2022}-2^x=2^{2026-16}\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

\(2VT=2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2022}\)

\(VT=2VT-VT=2^{x+2022}-2^x\)

\(\Rightarrow2^{x+2022}-2^x=2^{2026}-16\)

\(\Leftrightarrow2^{2022}.2^x-2^x=2^{2026}-2^4\)

\(\Leftrightarrow2^x\left(2^{2022}-1\right)=2^4\left(2^{2022}-1\right)\)

\(\Leftrightarrow2^x=2^4\Rightarrow x=4\)

\(2^x+2^{x+1}+2^{x+2}+...+2^{x+2021}=2^{2026}-16\)

\(\Rightarrow2^x\left(1+2^x+2^{x+1}+...+2^{x+2020}\right)=2^{2026}-2^4\)

\(\Rightarrow2^x.\dfrac{2^{x+2020+1}-1}{2-1}=2^4.\left(2^{2022}-1\right)\)

\(\Rightarrow2^x.\left(2^{x+2021}-1\right)=2^4.\left(2^{2022}-1\right)\)

\(\Rightarrow x\in\varnothing\)

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

b, 2\(x\) - 2016 = 2⁴.4

    2\(x\) - 2016  = 64

    2\(x\)               = 64 + 2016

    2\(x\)                = 2080

      \(x\)                = 2080 : 2

       \(x\)                = 1040

\(a,\Leftrightarrow6x-9+4-2x=-3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-3-6x\right)\left(2x-3+6x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-3-4x=0\\8x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{8}\end{matrix}\right.\)

Đặt biểu thức ở vế trái là A.

Ta có: \(A+3=\frac{2x+3y+4z+2022}{1+2x}+\frac{2x+3y+4z+2022}{1+3y}+\frac{2x+3y+4z+2022}{1+4z}=\frac{4038}{1+2x}+\frac{4038}{1+3y}+\frac{4038}{1+4z}\ge4038.\frac{9}{3+2x+3y+4z}=4038.\frac{9}{2019}=18\)

Dấu "=" xảy ra khi và chỉ khi 2x = 3y = 4z = 672

1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)

\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)

2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)