=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

k mk  đi , mk k lại cho

\(A=\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+....+\frac{3}{197\times200}\)

\(A=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+.....+\frac{1}{197\times200}\)

\(A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....\frac{1}{197}-\frac{1}{200}\)

\(A=\frac{1}{1}-\frac{1}{200}\)

\(A=\frac{199}{200}\)

a = 1/1- 1 /197 - 1 /200

a = 1/1 - 1/200

a = 199/200

\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{37\times40}\\ =\dfrac{1}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{37\times40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\dfrac{39}{40}\\ =\dfrac{13}{40}\)

=3/1-3/4+3/4-3/7+3/7-3/10+3/10-3/13

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

=1-1/13

=12/13

3/1x4+3/4x7+3/7x10+3/10x13

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

=1-1/13

=12/13

\(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+\frac{3}{10\times13}\)

\(=3\times\left(\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}\right)\)

\(=3\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)

\(=3\times\left(1-\frac{1}{13}\right)\)

\(=3\times\frac{12}{13}\)

\(=\frac{36}{13}\)

=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

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

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

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

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

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

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

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

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

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

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 !