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\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
x- 32:16=48
=>x-2=48
=>x=48+2
=>x=50
(x-32):16=48
=>x-32= 48x 16
=>x-32= 768
=>x=768+32
=>x=800
a) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=A=1-\frac{1}{2^{99}}\)
b) \(B=\frac{1}{2^2}+\frac{1}{2^4}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2^2.B=4B=1+\frac{1}{2^2}+......+\frac{1}{2^{98}}\)
\(\Rightarrow4B-B=3B=1-\frac{1}{2^{100}}\)
\(\Rightarrow b=\frac{1-\frac{1}{2^{100}}}{3}\)
a) Ta có A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
=> 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
=> A = \(1-\frac{1}{2^{99}}\)
b) Ta có B = \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)
=> 22B = \(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}=4B\)
=> 4B - B = \(\left(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\right)\)
=> 3B = \(1-\frac{1}{2^{100}}\)
=> B = \(\frac{1}{3}-\frac{1}{2^{100}.3}\)
Ta có : 5 : 4 dư 1 suy ra 5 -1 chia hết cho 4
5^2 :4 dư 1 suy ra 5^2 -1 chia hết cho 4
5^3 :4 dư 1 suy ra 5^3 -1 chia hết cho 4
suy ra 5^n : 4 dư 1 suy ra 5^n - 1 chia hết cho 4
Vậy 5^n - 1 chia hết cho 4 với n thuộc N
tk mk nha
5 : 4 dư 1 thì 5n với n thuộc Z chia cho 4 cũng dư 1
=> Vậy nếu 5n - 1 thì tất nhiên Chia hết cho 4
\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)
\(=2x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(2x^2+x+1\right)\)
Vậy a = 2; b = 1; c = 1.
(2011x2012+2012x2013)x(1+\(\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\))
= A x(1+\(\frac{1}{3}-1\frac{1}{3}\))
=A x(\(\frac{4}{3}-1\frac{1}{3}\))
= A x 0
=0
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{3}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{7}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{15}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{31}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{63}{64}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\Rightarrow64A=32+16+8+4+2+1\Rightarrow64A=63\Rightarrow A=\frac{63}{64}\)