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

= 1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

= 1 - 1/46 = 45/46 < 1

Cho S=3/1x4+3/4x7+3/7x10+...+3/40x43+3/43x46. Hãy chứng tỏ S<1

ĐPM : S < 1

S=3/1x4+3/4x7+3/7x10+...+3/40x43+3/43x46

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

\(S=1-\frac{1}{46}\)

=>S<1

S = 3/1.4 + 3/4.7 +....+ 3/43.46

S = 1 - 1/4 + 1/4 - 1/7 +.....+ 1/43 - 1/46

S = 1 - 1/46

S = 45/46 < 1

=> S < 1 (đpcm)

\(=\frac{1}{3}x\left(\frac{3}{1x4}+\frac{3}{4x7}+...+\frac{3}{77x80}\right)\)

\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{80}\right)\)

\(=\frac{1}{3}\times\frac{79.}{80}\)

\(=\frac{79}{240}\)

Tk giúp mk nha cảm ơn !!

\(=\frac{79}{240}\)

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

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

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

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

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

=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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)

\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)

\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)

Em cảm ơn chị