câu a 

quá dễ

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

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

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

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

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

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

Tk giúp mk nha cảm ơn !!

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

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{99.100}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

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 = \(\dfrac{93}{94}\)

=1−14 +14 −110 +...+119 −122 

=1−122 

 =  \(\frac{21}{22}\)

k cho mình nha chắc chắn đúng 100 %

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\)

\(=1-\frac{1}{22}\)

\(=\frac{21}{22}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{16.19}+\frac{3}{19.22}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\frac{19}{20}\)

\(=\frac{19}{60}\)

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

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