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?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bài 1:

$M=3.4.5+4.5.6+...+13.14.15$

$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$

$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$

$=-2.3.4.5+13.14.15.16=43560$

$M=43560:4=10890$

Bài 2:

a.

$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$

$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$

$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

$M=\frac{99}{100}:3=\frac{33}{100}$

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

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

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

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

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

Tk giúp mk nha cảm ơn !!

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

a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)

\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)

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

\(=2.\left(1-\dfrac{1}{100}\right)\)

\(=2.\dfrac{99}{100}\)

\(=\dfrac{99}{50}\)

_____

b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)

\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)

\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=3.\left(1-\dfrac{1}{101}\right)\)

\(=3.\dfrac{100}{101}\)

\(=\dfrac{300}{101}\)

dấu " . "  là dấu nhân nha bn, nãy viết nhanh quên lớp