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Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
Câu 8( Mình không viết đè nữa nha)
a) 2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100
= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100
= 1 – 1/100 < 1
= 99/100 < 1
Vậy A< 1
C=(1+1+1+...+1)+(1/1*3+1/2*4+1/3*5+...+1/2015*2017+1/2015*2017)
C=2015+(2/1*3+2/2*4+2/3*5+...+2/2015*2017+2/2015*2017):2
C=2015+(1-1/3+1/2-1/4+...+1/2015-1/2017+1/2015-1/2017):2
C=2015+(1+1/2-1/2016-1/2017+1/2015-1/2017)
cai nay thi ban tu tinh lay
nho k cho minh voi nhe
99.101 mới đúg nhé
=\(\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)
=\(\frac{2^2.3^2.4^2......100^2}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}=\frac{\left(2.3.4....100\right).\left(2.3.4....100\right)}{\left(1.2.3....99\right).\left(3.4.5......101\right)}\)
=\(\frac{100.2}{1.101}=\frac{200}{101}\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
Câu 1:
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...\left(1+\frac{1}{2014.2016}\right)\)
\(\Rightarrow C=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{2015.2015}{2014.2016}\)
\(\Rightarrow C=\frac{4.9.16...2015.2015}{3.8.15...2014.2016}\)
\(\Rightarrow C=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4...2014.2016}\)
\(\Rightarrow C=\frac{2.3.4...2015.2.3.4...2015}{1.2.3...2014.3.4.5...2016}\)
\(\Rightarrow C=\frac{2015}{1008}.\)
Vậy \(C=\frac{2015}{1008}.\)
Câu 2:
Do p là số nguyên tố lớn hơn 3 nên p có dạng \(3k+1\)hoặc\(3k+2\)
+ Nếu \(p=3k+1\Rightarrow p^2-1=\left(3k+1\right)^2-1\)
\(=9k^2+3k+3k+1-1\)
\(=9k^2+6k⋮3.\)( 1 )
+ Nếu \(p=3k+2\Rightarrow p^2-1=\left(3k+2\right)^2-1\)
\(=9k^2+6k+6k+4-1\)
\(=9k^2+12k+3⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow p^2-1⋮3\left(đpcm\right).\)
Câu 3:
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}>1000^{10}=10^{30}.\)( 1 )
\(2^{100}=2^{31}.2^6.2^{63}=2^{31}.64.512^7< 2^{31}.125.625^7=2^{31}.5^{31}=\)\(10^{31}.\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow10^{30}< 2^{100}< 10^{31}.\)
\(\Rightarrow\)2100 khi viết trong hệ thập phân có 31 chữ số.
Đáp số: 31 chữ số.
Câu 1 :
C = (1 + 1/1.3)(1 + 1/2.4)(1 + 1/3.5) .... (1 + 1/2014.2016)
C = (1.3/1.3 + 1/1.3) (2.4/2.4 + 1/2.4) ... (2014.2016/2014.2016 + 1/2014.2016)
C = 2.2/1.3 * 3.3/2.4 * ... * 2015.2015/2014.2016
C = 2.3....2015/1.2....2014 * 2.3....2015/3.4....2016
C = 2015 * 1/1008
C = 2015/1008