số các chữ số đó là

(200-1):1+1=200

số cặp đó là

200:2=100

tổng 1 cạp là

200+1=201

giá trị bt là 

201.100=20100

Ta có : \(S=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{1024}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+.....+\frac{1}{1024}\)

\(\Rightarrow2S-S=2-\frac{1}{1024}\)

\(\Rightarrow S=\frac{2047}{1024}\)

a: \(\dfrac{5}{13}\left(\dfrac{6}{29}-\dfrac{26}{39}\right)-\dfrac{6}{29}\cdot\left(\dfrac{5}{13}-\dfrac{29}{6}\right)\)

\(=\dfrac{5}{13}\cdot\dfrac{6}{29}-\dfrac{5}{13}\cdot\dfrac{26}{39}-\dfrac{6}{13}\cdot\dfrac{5}{13}+\dfrac{6}{29}\cdot\dfrac{29}{6}\)

\(=\dfrac{-5}{39}\cdot2+1=1-\dfrac{10}{39}=\dfrac{29}{39}\)

b: \(\dfrac{1\cdot198+2\cdot197+3\cdot196+...+198\cdot1}{1\cdot2+2\cdot3+...+198\cdot199}\)

\(=\dfrac{1\left(199-1\right)+2\left(199-2\right)+...+198\cdot\left(199-198\right)}{1\left(1+1\right)+2\left(1+2\right)+...+198\left(1+198\right)}\)

\(=\dfrac{199\left(1+2+...+198\right)-\left(1^2+2^2+...+198^2\right)}{\left(1+2+...+198\right)+\left(1^2+2^2+...+198^2\right)}\)

\(=\dfrac{199\cdot\dfrac{198\cdot199}{2}-\dfrac{198\cdot\left(198+1\right)\cdot\left(2\cdot198+1\right)}{6}}{198\cdot\dfrac{199}{2}+\dfrac{198\left(198+1\right)\left(2\cdot198+1\right)}{6}}\)

\(=\dfrac{3\cdot198\cdot199^2-198\cdot199\cdot397}{6}:\dfrac{3\cdot198\cdot199+198\cdot199\cdot397}{6}\)

\(=\dfrac{198\cdot199\left(3\cdot199-397\right)}{198\cdot199\left(3+397\right)}\)

\(=\dfrac{200}{400}=\dfrac{1}{2}\)

pls mai mình phải nộp rồi

Đầu tiên thì nhắc lại cái hằng đẳng thức cho bạn nào chưa học này: (a-b)²=a²-2ab+b²<=>a²+b²=(a-b)²+2ab

\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(=\dfrac{\left(\left(1-2\right)^2+2.1.2\right)}{1.2}+\dfrac{\left(\left(2-3\right)^2+2.2.3\right)}{2.3}+...+\dfrac{\left(\left(9-10\right)^2+2.9.10\right)}{9.10}\)
\(=\dfrac{\left(\left(-1\right)^2\right)}{1.2+2}+\dfrac{\left(\left(-1\right)^2\right)}{2.3+2}+...+\dfrac{\left(\left(-1^2\right)\right)}{9.10+2}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(=1-\dfrac{1}{10}+18\)
\(=18,9=\dfrac{189}{10}.\)

~ K chắc là đúng đâu ~

\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+...+\frac{181}{9.10}\)

=\(\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+...+\frac{180+1}{90}\)

=\(2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+...+2+\frac{1}{9.10}\)

=\(18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

=\(9-\frac{1}{10}\)

=\(\frac{189}{10}\)

\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+\frac{41}{4.5}+...+\frac{181}{9.10}\) \(=\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+\frac{40+1}{20}+...+\frac{180+1}{90}\) 

                                                                        \(=2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+2+\frac{1}{4.5}+...+2+\frac{1}{9.10}\) 

\(=18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=19-\frac{1}{10}\)

\(=\frac{189}{10}\)

a ta co ;

13 -12 +11+10-9+8-7-6+5-4+3+2-1

=13-(12-11-10+9) +(8-7-6+5) -(4-3-2+1)

= 13 -0+0 -0

=13

câu a = 13 còn câu b thì để tuần sau nhé

13-12+11+10-9+8-7-6+5-4+3+2-1

=11+11+10-9+8-7-6+5-4+3+2-1

=    22  +  1  + 1  - 11  - 7  + 1

=          23     +1  -11-7+1

=                24 -11-7+1

=                    13-8

=                        5

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)