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1/1024 câu này trên violimpic vòng 2 và mình làm đúng rồi
ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16};\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.....-\frac{1}{1024}\)
\(=1-\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{16}-\frac{1}{16}-....-\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
a) Ta có: \(\left(0.25\right)^4\cdot1024\)
\(=\left(0.25\right)^4\cdot4^4\cdot4\)
\(=\left(0.25\cdot4\right)^2\cdot4\)
\(=1^2\cdot4=4\)
b) Ta có: \(\frac{230^3}{23^3}\)
\(=\left(\frac{230}{23}\right)^3\)
\(=10^3=1000\)
c) Ta có: \(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)
\(=\left(-7\right)^n:\left[\frac{\left(-7\right)^n}{-7}\right]\)
\(=\left(-7\right)^n\cdot\frac{-7}{\left(-7\right)^n}\)
\(=-7\)
\(\left(\dfrac{1}{7}\right)^7\cdot7^7=\left(\dfrac{1}{7}\cdot7\right)^7=1^7=1\\ \left(0,125\right)^3\cdot512=\left(0,125\right)^3\cdot8^3=\left(0,125\cdot8\right)^3=1^3=1\\ \left(0,25\right)^4\cdot1024=\left(0,25\right)^4\cdot256\cdot4=\left(0,25\right)^4\cdot4^4\cdot4=\left(0,25\cdot4\right)^4\cdot4=1^4\cdot4=4\)
a) \(\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1^7=1\)
b) \(\left(0.125\right)^3.512=\left(0.125\right)^3.8^3=\left(0.125\cdot8\right)^3=1^3=1\)
c) \(\left(0.25\right)^4.1024=\left(0.25\right)^4.4^5=\left(0.25\right)^4.4^4.4=\left(0.25.4\right)^4.4=1^4.4=1.4=4\)
(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
A) (1/5)^5 . 5^5 = 1/5^5 . 5^5 = 5^5 / 5^5 = 1.
B)(0,125)^3 . 512 = (1/8)^3 . 512 = 1/8^3 . 512 = 1/512 . 512 = 1.
C) (0,25)^4 . 1024 = (1/4)^4 . 1024 = 1/4^4 . 1024 = 1/256 . 1024 = 4.
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Giả sử A = -B
\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=2-\frac{1}{1024}\)
\(B=\frac{2047}{1024}\)
=> \(A=-\frac{2047}{1024}\)
Ta có :
\(S=-1-\frac{1}{2}-\frac{1}{4}-......-\frac{1}{1024}\)
\(\Rightarrow-2S=2+1+\frac{1}{2}+.....+\frac{1}{512}\)
\(\Rightarrow-2S+S=\left(2+1+\frac{1}{2}+.....+\frac{1}{512}\right)+\left(-1-\frac{1}{2}-\frac{1}{4}-......-\frac{1}{1024}\right)\)
\(\Rightarrow-S=2-\frac{1}{1024}\)
\(\Rightarrow S=-2+\frac{1}{1024}\)
Đặt \(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\) và \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)
=>A=-B
\(B=1+\dfrac{1}{2}+...+\dfrac{1}{1024}\)
=>\(\dfrac{1}{2}B=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{11}}\)
=>\(-\dfrac{1}{2}B=\dfrac{1}{2^{11}}-1\)
=>\(\dfrac{1}{2}B=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)
=>\(B=\dfrac{2^{11}-1}{2^{10}}\)
=>\(A=\dfrac{1-2^{11}}{2^{10}}\)
(0,25)4.1024 = (0,25)4.256.4 = (0,25)4.44.4 = (0,25.4)4.4 = 1.4 = 4