Rút gọn A= 2^100-2^99+2^98-2^97+...+2^2-2
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Rút gon A =2^100-2^99+2^98-2^97+...+2^2-2
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+...+2^2-2\right)\)
\(\Rightarrow3A=2^{101}-2\)
\(\Rightarrow A=\frac{2^{101}-2}{3}\)
Vậy ......
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\(2A=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow2A+A=2^{101}-2\)
\(\Rightarrow3A=2^{101}-2\)
\(\Rightarrow A=\frac{2^{101}-2}{3}\)
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Rút gọn : a) M = \(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
b) N = \(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
a)M=2100-299+298-...+22-2
22M=2102-2101+2100-...+22-2
4M-M=2102-2101+2100-...+22-2-2100+299-...-22+2
3M=2102-2101
M=\(\frac{2^{102}-2^{101}}{3}\)
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Rút gọn biểu thức:
a/ A=100^2-99^2+98^2-97^2+...+2^2-1^2
b/ B=3(2^2+1)(2^4+1)...!2^64+1)+1
\(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)
\(=\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+....+\left(2-1\right).\left(2+1\right)\)
\(=1+2+....+97+98+99+100=\frac{100.\left(100+1\right)}{2}=5050\)
\(B=3\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)
Tiếp tục rút gọn như vậy,ta đc \(B=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1+1=2^{128}\)
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Rut gon
A=2^100-2^99+2^98-2^97+...+2^2+2
B=3^50-3^49+3^48-3^47+...+3^2-3+1
Tính:
a) A=2^100 - 2^99 + 2^98 - 2^97 + ... + 2^2 - 2
b) B=3^100 - 3^99 + 3^98 - 3^97 + ... + 3^2 - 3
thu gọn các tổng :
A=2^100 - 2^99 +2^98 - 2^97 +...+ 2^2 - 2
B= 3^100 - 3^99 + 3^98 - 3^97 +...+ 3^2 - 3 +1
A = 2100 - 299 + 298 - 297 +...+ 22 - 2
=> 2A = 2101 - 2100+299 - 298+...+23-22
=> 2A+A= 2101 -2
=> \(A=\frac{2^{101}-2}{3}\)
phần B bn lm tương tự nha!
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Thu gọn tổng sau:
a) A=1+3+3^2+...+3^100
b) B=2^100-2^99+2^98-2^97+...+2^2-2
c) C=3^100-3^99+3^98-3^97+...+3^2-3+1
a) A =1+3+32+33+...+3100
3A = 3 + 32+33+...+3101
3A-A=( 3 + 32+33+...+3101)-(1+3+32+33+...+3100)
2A = 3101-1
A = \(\frac{3^{101}-1}{2}\)
Thùy An làm sai rùi
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a) A=1+3+3^2+...+3^100
3A=3+3^2+....+3^101
3A-A=1+3^101
A=(1+3^101)/2
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a) A=1+3+32+...+3100
3A= 3+32+...+3100+3101
3A-A=3101-1
2A=3101-1
A=(3101-1):2
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Rút gọn:
a/ A=2^100-2^99+2^98-2^97+............+2^2-2
b/ B=3^100-3^99+3^98-3^97+..............+3^2-3+1
Ai nhanh nhất là đúng nhất mk tick cho
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
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Rút gọn:
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
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2^100-2^99+2^98-2^97+...+2^2-2
3^100-3^99+3^98-3^97+...+3^2-3+1
Giúp với các bạn ơi!!!
