tính :
C = \(\frac{6}{15.18}+\frac{6}{18.21}+.............+\frac{6}{87.90}\)
D = \(\frac{1}{25.27}+\frac{1}{27.29}+..............+\frac{1}{73.75}\)
E = \(\frac{3}{8.11}+\frac{3}{11.14}+................+\frac{3}{38.41}\)
help !!!!!!!!!!!!!
a) \(c=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
b) \(d=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
c) \(e=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
Mik đang cần gấp
Tính tổng dãy số cáh đều( tính nhanh)
a, \(\frac{7}{10.11}\)+ \(\frac{7}{11.12}\)+.....+\(\frac{7}{69.70}\)
b, \(\frac{6}{15.18}\)+ \(\frac{6}{18.21}\)+......+\(\frac{6}{87.60}\)
c, \(\frac{1}{25.27}\)+ \(\frac{1}{27.29}\)+.......+\(\frac{1}{73.75}\)
d, \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ \(\frac{1}{3.4.5}\)+.....+\(\frac{1}{18.19.20}\)
Làm chi tiết hộ mình nha
bài 3: chứng tỏ rằng:
b) Đặt A = \(\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}
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
D=\(\frac{6}{15.18}\)+\(\frac{6}{18.21}\)+\(\frac{6}{21.24}\)+.........+\(\frac{6}{87.90}\)
D=\(\frac{6}{15.18}\)+\(\frac{6}{18.21}\)+...+\(\frac{6}{87.90}\)
D=2.\(\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
D=2.\(\frac{1}{18}\)
D=\(\frac{1}{9}\)
Vậy D=\(^{\frac{1}{9}}\)
\(D=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(D=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(D=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(D=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(D=2.\left(\frac{6}{90}-\frac{1}{90}\right)\)
\(D=2.\frac{1}{18}\)
\(D=\frac{1}{9}\)
Tính hợp lí
\(A=\left(\frac{21}{31}+\frac{-16}{7}\right)+\left(\frac{44}{53}+\frac{10}{21}\right)+\frac{9}{53}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(A=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{21}+\frac{9}{53} \)
\(A=\left(\frac{16}{7}+\frac{10}{21}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)+\frac{21}{31}\)
\(A=\frac{58}{21}+1+\frac{21}{31}\)
\(A=\frac{100}{21}\)
\(B=6\left(\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=6.\frac{1}{18}\)
\(B=\frac{1}{3}\)
Tính:
\(A=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)
\(\Rightarrow2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)
\(\Rightarrow2A=\frac{1}{25}-\frac{1}{75}=\frac{3}{75}-\frac{1}{75}=\frac{2}{75}\)
\(\Rightarrow A=\frac{2}{75}\div2=\frac{1}{75}\)
tính hợp lí
\(\frac{6}{\text{15.18}}\)+\(\frac{6}{18.21}\)+\(\frac{6}{21.24}\)+...........+\(\frac{6}{87.90}\)
Ta có: \(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(=2\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(=2\cdot\frac{1}{18}=\frac{1}{9}\)
\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+.......+\frac{6}{87.90}\)
\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+.......+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(=2.\frac{1}{18}\)
\(=\frac{1}{9}\)
Tính tổng:
a) A = \(\frac{1}{2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ .... + \(\frac{1}{23.24}\)
b) B = \(\frac{6}{15.18}\)+ \(\frac{6}{18.21}\)+ \(\frac{6}{21.24}\)+ .... + \(\frac{6}{87.90}\)
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=\(\frac{11}{24}\)
b)\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(=3.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(=3.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(=3.\frac{5}{90}\)
\(=\frac{5}{30}\)
\(=\frac{1}{6}\)
Tính các tổng sau đây:
a)A=\(\frac{1}{25.27}\) +\(\frac{1}{27.29}\) +....+\(\frac{1}{73.75}\)
b)B=\(\frac{1}{8.11}\) +\(\frac{1}{11.14}\) +\(\frac{1}{14.17}\) +....+\(\frac{1}{197.200}\)
CÁC BẠN ƠI,GIÚP MIK VỚI.AI NHANH,LÀM ĐÚNG VÀ ĐỦ 2 Ý THÌ MIK TICK CHO NHA.Thank you very much.
\(a,A=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)
\(A=\frac{1}{2}\left[\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+...+\frac{2}{73\cdot75}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{75}\right]=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(b,B=\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+...+\frac{1}{197\cdot200}\)
\(3B=\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\)
\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)
\(3B=\frac{1}{8}-\frac{1}{200}\)
\(3B=\frac{3}{25}\)
\(B=\frac{3}{25}:3=\frac{1}{25}\)
#)Giải :
a, \(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(A=\frac{1}{25}-\frac{1}{75}\)
\(A=\frac{2}{75}\)
b, \(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
\(B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
\(B=\frac{1}{8}-\frac{1}{200}\)
\(B=\frac{3}{25}\)
#~Will~be~Pens~#
\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(2A=\frac{1}{25}-\frac{1}{75}\)
\(2A=\frac{2}{75}\)
\(A=\frac{2}{75}:2=\frac{1}{75}\)
\(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
\(3B=\frac{1}{8}-\frac{1}{200}\)
\(3B=\frac{3}{25}\)
\(B=\frac{3}{25}:3=\frac{1}{25}\)