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B=1.3+2.4+3.5+...+97.99+98.100B=1.3+2.4+3.5+...+97.99+98.100
B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)
B=1.2+1+2.3+2+....+97.98+97+98.99+98B=1.2+1+2.3+2+....+97.98+97+98.99+98
B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)
B=98.99.1003+98.992B=98.99.1003+98.992
B=323400+4851=328251B=323400+4851=328251
Số đó=1.3 + 2.4 + 3.5 +....+ 98.100
= 1(2+1) + 2.(3+1) + 3.(4+1) +...+ 98(99+1)
= 1.2 + 1 + 2.3 + 2 + 3.4 + 3+....+ 98.99 +98
= (1.2 + 2.3 + 3.4+....98.99) + (1+2+3+....+98)
=323400 + 4851=328251
\(A=1.3+3.5+5.7+...+97.99\)
\(\Rightarrow6A=1.3.6+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+97+99.\left(101-95\right)\)
\(\Rightarrow6A=1.3.6+3.5.7-1.3.5+5.7.9-3.5.7+...+97.99.101-95.97.99\)
\(\Rightarrow6A=1.3.6+97.99.101-1.3.5\)
\(\Rightarrow6A=3.\left(1+97.33.101\right)\)
\(\Rightarrow2A=1+323301\)
\(\Rightarrow2A=323302\)
\(\Rightarrow A=161651\)
A = 1( 2+1 ) + 2( 3+1 ) + 3( 4+1 ) +...+ 97( 98+1 ) + 98( 99+1 )
A = 1.2 + 1.1 + 2.3 + 2.1 + 3.4 + 3.1 +...+ 97.98 +97.1 + 98.99 + 98.1
A = ( 1.2 + 2.3 + 3.4 +...+ 97.98 + 98.99 ) + ( 1 + 2 + 3 +....+ 97 + 98)
A = 323400 + 4851 = 328251
\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
P = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/49.51
P = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/51
P = 1 - 1/51
P = 50/51
Q = 1/1.3 + 1/3.5 + ... + 1/19.21
Q = 1/2 .(2/1.3 + 2/3.5 + ... + 2/19.21)
Q = 1/2.(1 - 1/3 + 1/3 - 1/5 + ... + 1/19 - 1/21)
Q = 1/2 . (1 - 1/21)
Q = 1/2. 20/21
Q = 10/21
Ủng hộ mk nha ^_-
\(P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(P=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(P=1-\frac{1}{51}\)
\(P=\frac{50}{51}\)
\(Q=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(Q=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\right)\)
\(Q=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(Q=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(Q=\frac{1}{2}.\frac{20}{21}\)
\(Q=\frac{10}{21}\)
ta có:
C = 1 - 1/3 + 1/3 - 1/5 +...+1/69 - 1/71 + 1/71 - 1/73
= 1 - 1/ 73
= 72/73
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{69.71}+\)\(\frac{2}{71.73}\)
\(C=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{69}-\frac{1}{71}+\frac{1}{71}-\frac{1}{73}\)
\(C=1-\frac{1}{73}\)
\(C=\frac{72}{73}\)
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\)
\(=\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\right):2\)
\(=\left(2-\frac{2}{99}\right):2=\frac{98}{99}\)
D = 1 - 1/3 + 1/3 - 1/5 + .... + 1/97 - 1/99
D = 1 - 1/99
D = 98/99