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Mình cho đề bài thế này nhé \(2^x+2^{x+1}+2^{x+2}+...+2^{x+2017}=2^{2020}-4\) (1)
Nhân cả 2 vế của (1) cho 2, ta được \(2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2018}=2^{2021}-8\) (2)
Lấy (2) trừ theo vế với (1), ta thu được \(2^{x+2018}-2^x=2^{2020}-4\)
\(\Leftrightarrow2^x.2^{2018}-2^x=2^2.2^{2018}-2^2.1\)
\(\Leftrightarrow2^x\left(2^{2018}-1\right)=2^2\left(2^{2018}-1\right)\)
do \(2^{2018}-1\ne0\) nên ta hoàn toàn có thể suy ra \(2^x=2^2\Leftrightarrow x=2\)
Vậy \(x=2\)
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Ta có A = \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2021}\)
= \(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)
=> 2A - A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2021}}\right)\)
=> A = \(\frac{1}{2}-\frac{1}{2^{2021}}< \frac{1}{2}\left(\text{ĐPCM}\right)\)
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a: =5-78*32
=5-2496
=-2491
b: \(=6\left(9-6\right)=6\cdot3=18\)
c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)
d: \(=181+3-84+8\cdot25\)
=100+200
=300
e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)
=14000-1
=13999
f: \(=3^3+25\cdot8-1=26+200=226\)
g: \(=3+2^4+1=16+4=20\)
h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)
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S = 1 - 2 + 22 - 23 + ....... + 22020
2S = 2(1 - 2 + 22 - 23 + ....... + 22020)
2S = 2 - 22 + 23 - 24 + ....... + 22021
S = (2 - 22 + 23 - 24 + ....... + 22021) - (1 - 2 + 22 - 23 + ....... + 22020)
S = 22021 - 1
3S = 3(22021 - 1)
3S - 22021 = 3(22021 - 1) - 22021
3S - 22021 = 3.22021 - 3 - 22021
➤ 3S - 22021 = 22021 . 2 - 3
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Lời giải:
$2^2.85+15.2^2-2020^0=4.85+15.4-1$
$=4(85+15)-1=4.100-1=400-1=399$
22.85+15.22−20200=4.85+15.4−122.85+15.22−20200=4.85+15.4−1
=4(85+15)−1=4.100−1=400−1=399
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\(b, 2020^ 0 + 5^ 4 : 5^ 2 - 9.2 \)
\(=1+25-18\)
\(= 26-18=8\)
a)
(2.31.12)+(4.6.42)+(8.27.3)
=(24.31)+(24.42)+(24.27)
=24.(31+42+27)=24.100=2400
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A = 1 + 2 + 22 + 23 + ... + 22019
2A = 2(1 + 2 + 22 + ... + 22019)
2A = 2 + 22 + 23 + ... + 22020
=> 2A - A = (2 + 22 + 23 + ... + 22020) - (1 + 2 + 22 + .. +22019)
=> A = 22020 - 1
B - A = 22020 - (22020 - 1) = 1
\(A=1+2+2^2+2^3+...+2^{2019}\)
\(2A=2\left(1+2+2^2+2^3+...+2^{2019}\right)\)
\(2A=2+2^2+2^3+2^4+...+2^{2020}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2020}\right)-\left(1+2+2^2+2^3+...+2^{2019}\right)\)
\(A=2^{2020}-1\)
=> B - A = \(2^{2020}-\left(2^{2020}-1\right)=\text{ấn máy tính đel ra :))))}\)
Giải :
\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)
\(\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)
\(=2.2^2.3+...+2^{2020}.⋮3\)
Vậy \(S⋮3\)