1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100

= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)

= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)

= 1/3(1 - 1/100)

= 1/3*99/100

= 33/100

trả lời 

=33/100

chúc bn

học tốt

1.      \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}\)

\(=\frac{42}{43}\)

2.     Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=2.\left(1-\frac{1}{10}\right)\)

\(=2.\frac{9}{10}\)

\(=\frac{9}{5}\)

Ủng hộ mk nha !!! ^_^

1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43

= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43

= 1 - 1/43

= 42/43

2) 2/2 + 2/6 + 2/12 + ... + 2/90

= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)

= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)

= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)

= 2 × (1 - 1/10)

= 2 × 9/10

= 9/5

1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43

= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43

= 1 - 1/43

= 42/43

2) 2/2 + 2/6 + 2/12 + ... + 2/90

= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)

= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)

= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)

= 2 × (1 - 1/10)

= 2 × 9/10

= 9/5

S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103

S=1/1-1/103

S=102/103

Vì 102/103<1 nên S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{100\cdot103}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)

\(S=1-\frac{1}{103}\)

\(S=\frac{102}{103}< 1\)

\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+.......+\frac{3}{100x103}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}....+\frac{1}{100}-\frac{1}{103}\)

\(=\frac{1}{1}-\frac{1}{103}\)

=\(\frac{102}{103}\)

\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(\Leftrightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(\Leftrightarrow A=1-\frac{1}{46}\)

\(\Leftrightarrow A=\frac{45}{46}\)

Các bạn ơi. Chỗ cuối ko có số 4 đâu nha. Mình viết lộn

A = \(\frac{3}{1.4}\)\(+\)\(\frac{3}{4.7}\)\(+\)\(...\)\(+\)\(\frac{3}{43.46}\)

 = 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\) - \(\frac{1}{7}\)+ ...+\(\frac{1}{43}\)- \(\frac{1}{46}\)

= 1 - \(\frac{1}{46}\)

= \(\frac{45}{46}\)

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

D = \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.10}\)+...+ \(\dfrac{1}{91.94}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) -  \(\dfrac{1}{7}\) +  \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\)+...+ \(\dfrac{1}{91}\) - \(\dfrac{1}{94}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{94}\)

D = \(\dfrac{93}{94}\)

mk đang cần gấp

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

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

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

tại sao lại như thế?

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

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

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

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

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

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