a: \(A=\left(\frac{136}{15}-\frac{28}{5}+\frac{62}{10}\right)\cdot\frac{21}{24}\)

\(=\left(\frac{272}{30}-\frac{168}{30}+\frac{186}{30}\right)\cdot\frac78\)

\(=\frac{290}{30}\cdot\frac78=\frac{29}{3}\cdot\frac78=\frac{203}{24}\)

b: \(B=\frac56+6\frac56\left(11\frac{5}{20}-9\frac14\right):8\frac13\)

\(=\frac56+\frac{41}{6}\cdot\left(11+\frac14-9-\frac14\right):\frac{25}{3}=\frac56+\frac{41}{6}\cdot2\cdot\frac{3}{25}\)

\(=\frac56+\frac{41}{25}=\frac{125}{150}+\frac{246}{150}=\frac{371}{150}\)

c: \(C=1+3+6+\cdots+1225\)

\(=\frac12\left(2+6+12+\cdots+2450\right)=\frac12\cdot\left(1\cdot2+2\cdot3+\cdots+49\cdot50\right)\)

\(=\frac12\cdot\left\lbrack1\left(1+1\right)+2\left(2+1\right)+\cdots+49\left(49+1\right)\right\rbrack=\frac12\cdot\left\lbrack\left(1+2+\cdots+49\right)+\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot\frac{50}{2}+\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot25+49\cdot50\cdot\frac{99}{6}\right\rbrack=\frac12\cdot\left\lbrack49\cdot25+49\cdot25\cdot33\right\rbrack=\frac12\cdot49\cdot25\cdot\left(33+1\right)\)

\(=49\cdot25\cdot\frac{34}{2}=49\cdot25\cdot17=20825\)

D = -1-1/3-1/6-1/10-...-1/1225

Suy ra : D/2=-1/2-1/6-1/12-....-1/2450 
Mà 1/2=1/(1.2)=1-1/2; 1/6=1/(2.3)=1/2-1/3;...1/2450=1/(49.50)=... 
D/2= -(1-1/2)-(-1/2-1/3)-...-(1/49-1/50) 
D/2= -1+1/2-1/2+1/3-....-1/49+1/50 
D/2= -1+1/50=-49/50 
D=(-49/50).2=-98/50

k nha

giúp tôi

giúp tôi nhanh lên với

Google

A =  -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)

    = -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))

Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)

Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))

               = 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))

               = 2×_{\(\dfrac{24}{50}\)}

                   _{=  \(\dfrac{24}{25}\)}

      _{Thay B vào A ta có :}

   _{A = -1-\(\dfrac{24}{25}\)}

 _{=> A = \(\dfrac{-49}{25}\)}

 _{Cho mik một tick nhé thankss}

\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)

\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)

\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)

\(\dfrac{1}{2}B=\dfrac{-49}{50}\)

\(B=\dfrac{-49}{25}\)

\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)

\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)

\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

=-2*49/50

=-49/25

\(=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)

\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=-2\cdot\dfrac{49}{50}=\dfrac{-49}{25}\)

\(P=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{1225}\right)+\left(1-\dfrac{1}{1275}\right)\\ \Rightarrow\dfrac{P}{2}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)+\left(\dfrac{1}{2}-\dfrac{1}{12}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{2550}\right)\\ =\left(\dfrac{1}{2}-\dfrac{1}{2\cdot3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3\cdot4}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{50\cdot51}\right)\\ =\dfrac{1}{2}\cdot49-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\\ =\dfrac{49}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\\ =\dfrac{49}{2}-\dfrac{1}{2}+\dfrac{1}{51}=\dfrac{1225}{51}\\ \Rightarrow P=\dfrac{2450}{51}\)