\(A=\dfrac{3}{7x10}+\dfrac{3}{10x13}+\dfrac{3}{13x16}+...+\dfrac{3}{97x100}\)

\(A=3x\left(\dfrac{1}{7x10}+\dfrac{1}{10x13}+\dfrac{1}{13x16}+...+\dfrac{1}{97x100}\right)\)

\(A=3x\dfrac{1}{3}x\left(\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=3x\dfrac{1}{3}x\left(\dfrac{1}{7}-\dfrac{1}{100}\right)\)

\(A=1x\left(\dfrac{100}{700}-\dfrac{7}{700}\right)=\dfrac{93}{700}\)

=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16

=1-1/16=15/16

Ta có : \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{13}-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

= 15/16

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Cách giải thì làm theo bạn Lê Mạnh Tiến Đạt ý

câu a 

quá dễ

1/1.4+1/4.7+1/7.10+1/10.13+1/13.16

=1/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16)

=1/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)

=1/3.(1/1-1/16)

=1/3.(16/16-1/16)=1/3.15/16=5/16

3/1x4+3/4x7+...+3/97x100

=1-1/4+1/4-1/7+1/7-...+1/97-1/100

=1-1/100

=99/100

đây kô phải toán lớp 1

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(A=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{97\times100}\right)\)

\(A=3\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\times\left(1-\frac{1}{100}\right)\)

\(A=3\times\frac{99}{100}\)

\(A=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+......+\frac{3^2}{97.100}\)

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

Đặt \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

Ta có: \(S=\frac{3}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{97.100}\right)\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)

\(S=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow A=3.S=3.\frac{99}{100}=\frac{297}{100}\)