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 \(C=4.\frac{1}{18}\)
\(C=\frac{2}{9}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)
\(A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(A=1.\left(1-\frac{1}{2016}\right)\)
\(A=1.\frac{2015}{2016}\)
\(A=\frac{2015}{2016}\)
b) \(B=\frac{5}{10.11}+\frac{5}{11.12}+...+\frac{5}{69.70}\)
\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(B=5.\frac{3}{35}\)
\(B=\frac{3}{7}\)
c) \(C=\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\)
\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(C=4.\frac{1}{18}\)
\(C=\frac{2}{9}\)
