Đề có sai không bạn? Phạm Nguyễn Thục Anh

ta có :

M=\(\frac{1}{2.15}+\frac{1}{15.3}+\frac{1}{3.21}+....+\frac{6}{87.90}\)

\(\frac{1}{6}\)M=\(\frac{1}{6.2.15}+\frac{1}{15.3.6}+\frac{1}{6.3.21}+...+\frac{6}{6.87.90}\)

\(\frac{1}{6}\)M=\(\frac{1}{12.15}+\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\)

\(\frac{1}{6}\)M=\(\frac{1}{3}\left(\frac{3}{12.15}+\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

\(\frac{1}{6}\)M=\(\frac{1}{3}\)(\(\frac{1}{12}-\frac{1}{15}+\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\))

\(\frac{1}{6}\)M=\(\frac{1}{3}\left(\frac{1}{12}-\frac{1}{90}\right)\)

\(\frac{1}{6}\)M=\(\frac{1}{3}.\frac{13}{180}\)

\(\frac{1}{6}\)M=\(\frac{13}{540}\)

M=\(\frac{13}{540}:\frac{1}{6}\)

M=\(\frac{13}{90}\)

Vậy M=\(\frac{13}{90}\)

thôi , xin lỗi đã làm mất thời gian của mọi người nha

mình biết làm rùi

Mình nghĩ bài 1 đề sai .

bài 1 và bài 2 sai đề

a,A=\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{23.24}\)

A=\(\frac{1}{2}+\frac{2}{1}-\frac{1}{3}+\frac{3}{1}-\frac{1}{4}+......\frac{23}{1}-\frac{1}{24}\)

A=\(\frac{1}{2}-\frac{1}{24}\)

A=\(\frac{11}{24}\)

Còn câu b bạn??

b)\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

\(=3.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(=3.\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(=3.\frac{5}{90}\)

\(=\frac{5}{30}\)

\(=\frac{1}{6}\)

Ta có: \(A=\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\)

                \(=\frac{1}{3}(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90})\)

                \(=\frac{1}{3}(\frac{1}{15}-\frac{1}{90})\)

                \(=\frac{1}{3}(\frac{6}{90}-\frac{1}{90})\)

                \(=\frac{1}{3}.\frac{5}{90}\)

                \(=\frac{1}{54}\)

Ta có: 1= \(\frac{54}{54}\)

Suy ra A < 1 (đpcm)

3A=3*(1/15*18+1/18*21+...+1/87*90)

3A=3/15*18+3/18*21+...+3/87*90

3A=1/15-1/18+1/18-1/21+...+1/87-1/90

3A=1/15-1/90

3A=1/18

A=1/18 chia3

A=1/54

vì 1/54<1 nên A<1

3A là gì vậy

Giải:

a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)

C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)

C = \(\frac{6}{3}.\frac{1}{18}\)

C = \(2.\frac{1}{18}\)

C = \(\frac{1}{9}\)

Vậy C = \(\frac{1}{9}\)

b) D = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

D = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)\

D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

D = \(\frac{1}{2}.\frac{2}{75}\)

D = \(\frac{1}{75}\)

Vậy D = \(\frac{1}{75}\)

c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)

E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)

E = \(\frac{1}{8}-\frac{1}{41}\)

E = \(\frac{33}{328}\)

Vậy E = \(\frac{33}{328}\)