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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=\(^{\frac{1}{9}}\)
\(D=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(D=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(D=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(D=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(D=2.\left(\frac{6}{90}-\frac{1}{90}\right)\)
\(D=2.\frac{1}{18}\)
\(D=\frac{1}{9}\)
\(M=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+.....+\frac{6}{87.90}\)
\(\Rightarrow M=6\left(\frac{1}{15.18}+\frac{1}{18.21}+\frac{1}{21.24}+....+\frac{1}{87.90}\right)\)
\(\Rightarrow M=6\left[\frac{1}{3}\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+.....+\frac{1}{87}-\frac{1}{90}\right)\right]\)
\(\Rightarrow M=6\left[\frac{1}{3}\left(\frac{1}{15}-\frac{1}{90}\right)\right]\Rightarrow M=6\left(\frac{1}{3}.\frac{1}{18}\right)\Rightarrow M=6.\frac{1}{54}\Rightarrow M=\frac{1}{9}\)
\(A=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{21}+\frac{9}{53} \)
\(A=\left(\frac{16}{7}+\frac{10}{21}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)+\frac{21}{31}\)
\(A=\frac{58}{21}+1+\frac{21}{31}\)
\(A=\frac{100}{21}\)
\(B=6\left(\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=6.\frac{1}{18}\)
\(B=\frac{1}{3}\)
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{4}{4}\left(\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=\frac{3}{3}\left(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\right)\)
\(B=2\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}=\frac{1}{9}\)
Trả lời:
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}\)
\(A=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}\)
\(B=\frac{1}{9}\)
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 = \(\frac{33}{328}\)
Mình nói lí thuyết cho nghe:
Với phân số \(\frac{a-b}{a.b}\)\(\left(VD:\frac{1}{1.2};\frac{1}{2.3};\frac{1}{2015.2016};\frac{3}{15.18};\frac{3}{18.21};\frac{1}{10.11};\frac{1}{11.12};...\right)\)thì:
\(\frac{b-a}{a.b}=\frac{b}{a.b}-\frac{a}{a.b}=\frac{1}{a}-\frac{1}{b}\left(VD:\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{3}{15.18}=\frac{1}{15}-\frac{1}{18}\right)\)
ÁP dụng để tính:
c) \(\Rightarrow\frac{1}{4}C=\frac{1}{4}\left(\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\right)=\frac{3}{15.18}+\frac{3}{18.21}+....+\frac{3}{87.90}\)
\(\Rightarrow\frac{1}{4}C=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}=\frac{1}{15}-\frac{1}{90}\)
=> \(C=\left(\frac{1}{15}-\frac{1}{90}\right).4\)
a,\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=1-\frac{1}{2016}\)suy ra \(A=\frac{2015}{2016}\)
b, \(B=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(B=5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(B=5\left(\frac{1}{10}-\frac{1}{70}\right)\)suy ra \(B=5.\frac{3}{35}\)
\(B=\frac{3}{7}\)
c,\(C=4.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)suy ra \(C=4.\frac{1}{18}\)
\(C=\frac{2}{9}\)
\(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\)
\(=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{84}-\frac{1}{87}+\frac{1}{87}-\frac{1}{90}\)
\(=\frac{1}{15}-\frac{1}{90}\)
\(=\frac{6}{90}-\frac{1}{90}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
1/15-1/18+1/18-1/21+1/21-1/24+....+1/87-1/90
=1/15-1/90
=6/90-1/90
=5/90
=1/16
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??