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\(\frac{561}{704}.\frac{39}{47}+\frac{561}{704}.\frac{8}{47}\)
\(=\frac{561}{704}.\left(\frac{39}{47}+\frac{8}{47}\right)\)
\(=\frac{561}{704}.1\)
\(=\frac{51}{64}\)
ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
Đặt tổng : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}=A\)
Ta tính : A x 2 - A = ( \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)) - (\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\) )
=\(1-\frac{1}{512}=\frac{511}{512}\)
Mà A x 2 - A = A Vậy A=\(\frac{511}{512}\)
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
a/ \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
= \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{256}-\frac{1}{512}\right)+\left(\frac{1}{512}-\frac{1}{1024}\right)\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)
= \(1-\frac{1}{1024}\)
= \(\frac{1023}{1024}\)
b/ \(\frac{1}{8}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{10200}\)
= \(\frac{1}{8}+\frac{1}{6\times8}+\frac{1}{8\times10}+...+\frac{1}{100\times102}\)
= \(\frac{1}{8}+\frac{1}{2}\times\left(\frac{2}{6\times8}+\frac{2}{8\times10}+...+\frac{2}{100\times102}\right)\)
= \(\frac{1}{8}+\frac{1}{2}\times\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{102}\right)\)
= \(\frac{1}{8}+\frac{1}{2}\times\left(\frac{1}{6}-\frac{1}{102}\right)\)
= \(\frac{1}{8}+\frac{1}{2}\times\frac{8}{51}\)
= \(\frac{1}{8}+\frac{4}{51}\)
= \(\frac{83}{408}\)
a ,Ta có : 14/25 < 15/25 = 3/5
Ta có : 1 - 3/5 = 2/5
1 - 5/7 = 2/7
Nên 2/5 > 2/7
Vậy 14/25 < 5/7
512-47=465
TL ;
512 - yx5 = 45
yx5 = yx = 512 - 45
yx5 = yx = 467
x = 465 - 5 = 460
y = 460 : 2 = 230