\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

Các bạn giải giúp mình nha😐

a: =-13/24

b: A

a) a có sai đề bài không

b) chọn A

a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

      \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

       \(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

                  \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

                   \(=3.\left(1-\frac{1}{100}\right)\)

                     \(=3.\frac{99}{100}=\frac{297}{100}\)

dễ ẹc

Ta có :

3-7=11-15=...=99-103

Từ 3 đến 103 có số số hạng là :

( 103 - 3 ) : 4 + 1 = 26 ( số )

Từ 3 đến 103 có số cặp là :

26 : 2 = 13 ( cặp )

Vậy1+ 3-7+11-15+19-23+...+99-103 = 13 . (-4) +1 = -51

Đáp Số : -51 

\(=\dfrac{1}{2}-\left(\dfrac{1}{3\cdot7}+\dfrac{1}{7\cdot11}+...+\dfrac{1}{23\cdot27}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{9-1}{27}\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{8}{27}=\dfrac{1}{2}-\dfrac{2}{27}=\dfrac{27-4}{54}=\dfrac{23}{54}\)

a) \(\frac{7}{18}+\frac{4}{23}+\frac{11}{18}+\frac{19}{23}\)

\(=\left(\frac{7}{18}+\frac{11}{18}\right)+\left(\frac{4}{23}+\frac{19}{23}\right)\)

\(=1+1\)

\(=2\)

b) \(\frac{19}{15}+\frac{31}{23}-\frac{4}{15}-\frac{8}{23}\)

\(=\left(\frac{19}{15}-\frac{4}{15}\right)+\left(\frac{31}{23}-\frac{8}{23}\right)\)

\(=1+1\)

\(=2\)

ủng hộ tớ nha

a) (7/18+11/18)+(4/23+19/23)

=18/18+23/23

=1+1

=2

b)=2

\(a,b=2\)

\(a,2011+2010\left(-4.5^2+11.3^2\right)^{2009}\)

\(=2011+2010\left(-1\right)=2011-2010\)

\(=1\)

A.5/7

B.-481/391

C.0

D.-59/9

E.1

\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)

=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)

= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)

= \(1\dfrac{11}{25}+\dfrac{5}{7}\)

= \(2\dfrac{27}{175}\)

b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)

=\(-2+\left(\dfrac{ }{ }\right)\)