Số cuối có trừ 2 koe

\(M=\left[\frac{1}{2}-1\right]\cdot\left[\frac{1}{3}-1\right]\cdot\left[\frac{1}{4}-1\right]\cdot...\cdot\left[\frac{1}{2011}-1\right]\)

\(M=\left[\frac{1}{2}-\frac{2}{2}\right]\cdot\left[\frac{1}{3}-\frac{3}{3}\right]\cdot\left[\frac{1}{4}-\frac{4}{4}\right]\cdot...\cdot\left[\frac{1}{2011}-\frac{2011}{2011}\right]\)

\(M=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\frac{-3}{4}\cdot...\cdot\frac{-2010}{2011}\)

\(M=\frac{\left[-1\right]\cdot\left[-2\right]\cdot\left[-3\right]\cdot...\cdot\left[-2010\right]}{2\cdot3\cdot4\cdot...\cdot2011}\)

\(M=\frac{1\cdot2\cdot3\cdot...\cdot2010}{2\cdot3\cdot4\cdot...\cdot2011}=\frac{1}{2011}\)

đụ cha mi

mi trù ta thi rớt HK II mà ta giúp mày hả

mấy bài này cũng dễ ẹt nữa

đừng có mơ ta sẽ giúp mày

ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
 

\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)

\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)

\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)

\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)

\(B=\frac{100\cdot2}{1\cdot101}\)

\(B=\frac{200}{101}\)

XD: best tiếng anh chuyển sang toán ak!?

\(B1:\)

\(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)

\(=\frac{16}{9}\cdot\frac{27}{20}\cdot\frac{40}{33}\cdot\cdot\cdot\frac{10807}{10800}\)

\(=\frac{8.2}{9.1}\cdot\frac{9.3}{10.2}\cdot\frac{10.4}{11.3}\cdot\cdot\cdot\frac{57.51}{58.50}\)

\(=\frac{\left(8.9.10...57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right).\left(1.2.3...50\right)}\)

\(=\frac{8.51}{58.1}=\frac{204}{29}\)

Vậy.....

\(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)

\(M=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{10807}{10800}\)

\(M=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}...\frac{107.101}{108.100}\)

\(M=\frac{\left(2.3.4...101\right)\left(8.9.10...107\right)}{\left(1.2.3...100\right)\left(9.10.11...108\right)}\)

\(M=\frac{101.8}{108}\)

\(M=\frac{202}{27}\)

k mình nha . câu 2 tí nữa mình gửi

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

đổi k ko,mk hứa sẽ k lại(nếu ko làm chó!!!!!!!!!!!!!)

Bài 1: <Cho là câu a đi>:

a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\) 

\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\) 

\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\) 

\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\) 

\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\) 

\(\rightarrow x+1=50\rightarrow x=49\) 

Vậy x = 49.

1.\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{49}{50}=\frac{1}{50}\)

1/ \(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{49}{50}=\frac{1.\left(2.3.4...49\right)}{50.\left(2.3.4....49\right)}=\frac{1}{50}\)

2/ Chưa học dạng này:v