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ta co:
\(\frac{11.12+22.24+44.48}{33.36+66.72+132.144}\)
\(=\frac{11.12+11.2.12.2+11.4.12.4}{33.36+33.2.36.2+33.4.36.4}\)
\(=\frac{11.12.\left(1+2.2+4.4\right)}{33.36.\left(1+2.2+4.4\right)}\)
\(=\frac{11.12}{36.33}\)
\(=\frac{1}{3.3}\)
\(=\frac{1}{9}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)

\(=7\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7\times\frac{6}{70}\)
\(=\frac{6}{10}=\frac{3}{5}\)
Gọi tổng trên là A
A = 7/10.11 + 7/11.12 +.....+ 7/69.70
A = 7(1/10.11 + 1/11.12 +.....+ 1/69.70)
A =7( 1/10 - 1/11 + 1/11 - 1/12 +.....+ 1/69 - 1/70)
A = 7( 1/10 - 1/70)
A = 7 . 3/35
A = 21/35

\(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+.......+\frac{1}{19\cdot20}\)\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{18}-\frac{1}{19}\)\(+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)
K CHO MÌNH NHA CÁC BẠN

Ta có : \(\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}=7\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)=7\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{6}{70}=\frac{3}{5}\)

\(A=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)
\(A=7\left(\dfrac{3}{35}\right)\)
\(A=\dfrac{3}{5}\)
Cách giải vậy đó