Tính
A= \(2012^{\left(256-1^2\right)\left(256-2^2\right)...\left(256-50^2\right)}\)
K
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MC
3 tháng 6 2018
A = ( 2 + 1 )( 22 + 1 )...( 2256 + 1 ) + 1
A = ( 2 - 1 )( 2 + 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 ) + 1
A = ( 22 - 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 ) + 1
A =( 24 - 1 ) ( 24 + 1 )...( 2256 + 1 ) + 1
A = ( 2256 - 1 )( 2256 + 1 ) + 1
A = 2512
HQ
Hà Quang Minh
Giáo viên
22 tháng 8 2023
\(a,\left(\dfrac{1}{256}\right)^{-0,75}+\left(\dfrac{1}{27}\right)^{-\dfrac{4}{3}}\\ =256^{\dfrac{3}{4}}+27^{\dfrac{4}{3}}\\ =\sqrt[4]{256^3}+\sqrt[3]{27^4}\\ =145\\ b,\left(\dfrac{1}{49}\right)^{-1,5}-\left(\dfrac{1}{256}\right)^{-\dfrac{2}{3}}\\ =49^{\dfrac{3}{2}}-256^{\dfrac{2}{3}}\\ \simeq343-40,3\\ \simeq302,7\)
6 tháng 8 2019
a)\(\left(2+1\right)\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
Tiếp tục như thế, ta được:
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)-1=2^{512}-1-1=2^{512}-2\)
b) \(24\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
Tiếp tục như thế, ta được:
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}=5^{64}-1-5^{64}=-1\)
HN
6 tháng 8 2019
\(\left(2+1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(\left(2-1\right).\left(2+1\right).\left(2^2+1\right).....\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right).\left(2^{256}+1\right)+1=2^{512}+1\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 12 2018
\(A-1=\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^2-1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^4-1\right)\left(x^4+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^{256}-1\right)\left(x^{256}+1\right)=x^{512}-1\)
\(\Rightarrow A-1=\dfrac{x^{512}-1}{x-1}\)
\(\Rightarrow A=\dfrac{x^{512}-1}{x-1}+1=\dfrac{x^{512}+x-2}{x-1}\)
12 tháng 1 2019
a) Đề sai nha bạn :) mấy dấu cộng bạn phỉa chuyển thành dấu nhân nhé
\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(A=2^{512}-1+1\)
\(A=2^{512}\)
12 tháng 1 2019
b . ( 5x - 3y + 4z )( 5x - 3y - 4z ) = ( 5x - 3y )^2 - ( 4z )^2 = 25x^2 - 30xy + 9y^2 - 16z^2 = 25( y^2 + z^2 ) - 30xy + 9y^2 - 16z^2 = 9z^2 + 34y^2 - 30xy ( 1 )
( 3x - 5y )^2 = 9x^2 - 30xy + 25y^2 = 9( y^2 + z^2 ) - 30xy + 25y^2 = 34y^2 + 9z^2 - 30xy ( 2 )
Tu ( 1 ) va ( 2 ) => dpcm
LH
2
H
14 tháng 10 2023
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\\ \Rightarrow\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{2}{15}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{6}{15}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{8}{15}\\ \left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\\ \Rightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\\ \Rightarrow2x+5=4\\ \Rightarrow2x=4-5\\ \Rightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
ND
Nguyễn Đăng Nhân
CTVHS
14 tháng 10 2023
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(x-\dfrac{2}{15}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}+\dfrac{2}{15}\)
\(x=\dfrac{8}{15}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
\(2x+5=4\)
\(2x=-1\)
\(x=-0,5\)