A=2 mũ 0 + 2 mũ 1 + 2 mũ 2 + ... + 2 mũ 2010
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\(A=2^0+2^1+2^2+...+2^{2010}\)
\(2A=2^1+2^2+2^3+...+2^{2021}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2021}\right)-\left(2^0+2^1+2^2+...+2^{2020}\right)\)
\(A=2^{2021}-1\)
\(A=2^0+2^1+2^2+2^3+...+2^{2010}\)
\(A=1+2+2^2+2^3+...+2^{2010}\)
\(2A=2+2^2+2^3+...+2^{2011}\)
\(2A-A=\left[2+2^2+2^3+...+2^{2011}\right]-\left[1+2+2^2+2^3+...+2^{2010}\right]\)
\(A=2^{2011}-1\)
Mà \(B=2^{2011}-1\)
=> A = B
Ta có: A=\(2^0+2^1+2^2+2^3+...+2^{2010}\)
2A=\(2^1+2^2+2^3+2^4+...+2^{2011}\)
2A-A hay A=\(2^{2011}-2^0\)
=\(2^{2011}-1\)
Vì \(2^{2011}-1=2^{2011}-1\)
\(\Rightarrow\)A=B
Hok tốt nha!!!
a) A = (7-9)3+ (-5)7 :(-5)5 +20100=(-2)3 + (-5)7-5 +1
=-8 +(-5)2 +1 = -8 +25 +1=18
B) b=(2.(-2)2 +1:(-3)2) +9.7 -(-2)6 =2.4 +1/9 + 63 - 64 = 8 +1/9 -1 =64/9
CHÚC BN GIỎI NHÉ <3
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
Câu 1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{99}=\frac{49}{100}\)
cho mình nha bạn
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
2A=2\(\left(2^0+2^1+2^2+...+2^{2010}\right)\)
2A=\(2^1+2^2+2^3+...+2^{2011}\)
2A-A=\(2^{2011}-2^0\)
Vậy A=\(2^{2011}-2\)
\(A=2^0+2 ^1+...+2^{2010}\\ \Rightarrow2.A=2+2^2+....+2^{2011}\\ \Rightarrow2.A-A=2^{2011}-1\\ \Rightarrow A=2^{2011}-1\)