=1/3(3/2*5+3/5*8+...+3/95*98)

=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)

=1/3*48/98

=1/3*24/49

=8/49

nay tui con thay

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{1}{3}\cdot\frac{99}{202}=\frac{33}{202}\)

sửa lại đề 1 chút : 

\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{98\cdot101}\)

\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{98\cdot101}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{1}{3}\cdot\frac{99}{202}\)

\(=\frac{33}{202}\)

\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)

\(\frac{A}{3}=\frac{23-20}{20.23}+\frac{26-23}{23.26}+...+\frac{80-77}{77.80}\)

\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\Rightarrow A=\frac{9}{80}< 1\)

3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21

=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21

=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21

x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21

x . ( 1/2 - 1/14 ) = 1/21

x . 3/7 = 1/21 

x = 1/21 : 3/7

=> x = 1/9

\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)

<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)

<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)

<=> \(x=\frac{1}{9}\)

Đặt vế trái là B

\(3B=\frac{23-20}{20.23}+\frac{26-23}{23.26}+\frac{29-26}{26.29}+...+\frac{80-77}{77.80}\)

\(3B=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}\)

\(3B=\frac{3}{80}\Rightarrow B=\frac{1}{80}< \frac{1}{9}\)

Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)

\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)

Vậy \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)

:3 B<1/79 mà 1/79<1/9=>B<1/9

= 1/3.(1/2-1/5)+1/3.(1/5-1/8)+....+1/3.(1/92-1/95)+1/3.(1/95-1/98)

=1/3.(1/2-1/5+1/5-1/8+....+1/92-1/95+1/95-1/98)

=1/3.(1/2-1/98)

=1/3.24/49

=8/49

Phân tích: 1/2.5 = 1/2 - 1/5

1/5.8 = 1/5 - 1/8

1/8.11 = 1/8 - 1/11

...

1/92.95 = 1/92 - 1/95

1/95.98 = 1/95 - 1/98

Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98

3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98

3 =  1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98

= 1 - 1/98

= 97/98 : 3 = 97/98 x 1/3 = (tự tính)

1/2.5+...+1/95.98

=1/2-1/5+1/5-1/8+....1/95+1/98

=1/2-1/98

=24/49

K NHA!!!

S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)

   = 1/3 . (1/2 - 1/20)

   = 1/3 . 9/20

   = 3/20

\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)

\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(S=\frac{9}{20}:3=\frac{3}{20}\)

\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...........+\frac{3}{17.20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{10}{20}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\frac{9}{20}\)

\(-\frac{3}{20}\)

a/

\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+66.67+66.2=\)

\(=\left(1.2+2.3+3.4+...+66.67\right)+2\left(1+2+3+...+66\right)\)

Đặt

\(B=1+2+3+...+66=\dfrac{66\left(1+66\right)}{2}=2211\)

Đặt 

\(C=1.2+2.3+3.4+...+66.67\)

\(3C=1.2.3+2.3.3+3.4.3+...+66.67.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+66.67.\left(68-65\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-65.66.67+66.67.68=\)

\(=66.67.68\Rightarrow C=\dfrac{66.67.68}{3}=22.67.68\)

\(\Rightarrow A=C+2B\) Bạn tự tính nhé

b/

\(B=2\left(1.50+2.49+3.48+...+25.26\right)=\)

Ta có

\(C=1.50+2.49+3.48+...+25.26=\)

\(\left(50-49\right).50+\left(50-48\right).49+\left(50-47\right).48+...+\left(50-25\right).26=\)

\(=50.50-49.50+50.49-48.49+50.48-47.48+50.26-25.26=\)

\(=50.\left(26+27+28+...+50\right)-\left(25.26+26.27+27.28+...+49.50\right)\)

Ta có

\(D=26+27+28+...+50=\dfrac{25.\left(26+50\right)}{2}=950\)

Ta có

\(E=25.26+26.27+27.28+...+49.50\)

\(3E=25.26.3+26.27.3+27.28.3+...+49.50.3=\)

\(=25.26.\left(27-24\right)+26.27.\left(28-25\right)+...+49.50.\left(51-48\right)=\)

\(=-24.25.26+25.26.27-25.26.27+26.27.28-...-48.49.50+49.50.51=\)

\(=49.50.51-24.25.26\)

\(\Rightarrow E=\dfrac{49.50.51-24.25.26}{3}\)

\(\Rightarrow C=50D-E\)

\(B=2C\)

Bạn tự tính nhé