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17 tháng 7 2019

3.5.17.257.65537=4294901760

11 tháng 7 2021

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)

11 tháng 7 2021

mk cảm ơn ah

 

13 tháng 9 2015

c;=(50-49)(50+49)+(48-47)(48+47)+.............+(2+1)(2-1)


=50+49+48+............+1


=(50+1)50=2550:2=1275


d;=(2^4-1)(2^4+1)(2^8+1)(2^16+1)


=(2^8-1)(2^8+1)(2^16+1)


=(2^16-1)(2^16+1)


=2^32-1



e;=(3-1)(3+1)(3^2+1)...........(3^16+1)


=(3^2-1)(3^2+1)..............(3^16+1)


=(3^16-1)(3^16+1)=3^32-1


tu tinh ket qua luy thua tao khong thua hoi dau



6 tháng 2 2016

= 36 đó bn

tich nha!!

23 tháng 2 2021

1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744

14 tháng 11 2025

Bài 1:

\(\frac14+\frac23=\frac{3}{12}+\frac{8}{12}=\frac{3+8}{12}=\frac{11}{12}\)

\(\frac27+\frac23=\frac{6}{21}+\frac{14}{21}=\frac{6+14}{21}=\frac{20}{21}\)

\(\frac25+\frac13=\frac{6}{15}+\frac{5}{15}=\frac{6+5}{15}=\frac{11}{15}\)

\(\frac23+\frac12=\frac46+\frac36=\frac{4+3}{6}=\frac76\)

\(\frac13+\frac35=\frac{5}{15}+\frac{9}{15}=\frac{5+9}{15}=\frac{14}{15}\)
\(\frac45+\frac13=\frac{12}{15}+\frac{5}{15}=\frac{12+5}{15}=\frac{17}{15}\)

\(\frac18+\frac34=\frac18+\frac68=\frac{1+6}{8}=\frac78\)

\(\frac{1}{36}+\frac{5}{12}=\frac{1}{36}+\frac{15}{36}=\frac{1+15}{36}=\frac{16}{36}=\frac49\)

\(\frac13+\frac16+\frac{1}{18}=\frac{6}{18}+\frac{3}{18}+\frac{1}{18}=\frac{6+3+1}{18}=\frac{10}{18}=\frac59\)

Bài 2:

\(\frac{15}{16}-\frac{3}{16}=\frac{15-3}{16}=\frac{12}{16}=\frac34\)

\(\frac{17}{18}-\frac56=\frac{17}{18}-\frac{15}{18}=\frac{17-15}{18}=\frac{2}{18}=\frac19\)

\(\frac34-\frac49=\frac{27}{36}-\frac{16}{36}=\frac{27-16}{36}=\frac{11}{36}\)

\(\frac12-\frac25=\frac{5}{10}-\frac{4}{10}=\frac{5-4}{10}=\frac{1}{10}\)

\(\frac56-\frac{3}{10}=\frac{25}{30}-\frac{9}{30}=\frac{25-9}{30}=\frac{16}{30}=\frac{8}{15}\)

\(3-\frac13=\frac93-\frac13=\frac{9-1}{3}=\frac83\)

\(\frac45-\frac{1}{10}=\frac{8}{10}-\frac{1}{10}=\frac{7}{10}\)

\(\frac52-1=\frac{5-2}{2}=\frac32\)

\(\frac58-\frac25=\frac{25}{40}-\frac{16}{40}=\frac{25-16}{40}=\frac{9}{40}\)

20 tháng 4 2016

b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)

                 \(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)

                 \(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)

                 \(=1+1,5+2+...+8,5\)

                 \(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)

19 tháng 4 2016

B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)                                                                               

 B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)

Vậy 1/8<2 hay 1/8<16/8

21 tháng 4 2016

A = 1/2^2 + 1/3^2 +.. + 1/8^2 < 1/1.2 + 1/2.3  +... + 1/7.8 = 1 - 1/2 + 1/2 -1/3 +...  + 1/7 - 1/8

=  1 - 1/8 < 1 

\(\Rightarrowđpcm\)

\(tíchnhaminhftchlaij\)