mà mình không phải là Thiên Tỷ mình tên là Gia Yến đấy nhé

=1-1/4+1/4-1/7+1/7-...+1/37-1/40

=1-1/40=39/40

Mình cần gấp ạ. Mốt mik thi rồi

Mình đang gấp lắm ạ

=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

Bài này giống toán lớp 6 hơn

m = 3/(1x4) + 3/(4x7) +  ... + 3/(19x22)

    = (4-1)/(1x4) + (7-4)/(4x7) +  ... + (22-19)/(19x22)

    = 4/(1x4) - 1/(1x4) + 7/(4x7) - 4/(4x7) + ... +  22/(19x22) - 19/(19x22)

    = 1 - 1/4 + 1/4 - 1/7 + ... + 1/19 - 1/22

    = 1-1/22

    = 21/22

A=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2.\left(x+3\right)}\)

=> A=\(\frac{3}{1}-\frac{3}{4}+\frac{3}{4}+...+\frac{3}{2.x}-\frac{3}{2.\left(x+3\right)}\) 

=> A =\(\frac{3}{1}-\frac{3}{2.\left(x+3\right)}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)

\(=1-\frac{1}{97}\)

\(=\frac{96}{97}\)

Bài làm:

Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)

\(=1-\frac{1}{97}\)

\(=\frac{96}{97}\)

Ta có \(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{7}=1-\frac{1}{7}=\frac{6}{7}\)

P/S : Dấu "." là dấu "x" nha !

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

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

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

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

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

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

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

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

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

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

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

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