(1/1.3+1/3.5+…+1/97.99).x=1/99
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Ta viết lại tổng này thành:
\(P=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{98.100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}-\dfrac{49}{99}\right)\)
\(P=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}-\dfrac{1}{198}+\dfrac{1}{4}-\dfrac{1}{200}-\dfrac{49}{99}\)
\(P=\dfrac{49}{200}\)

\(S=\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{97\cdot99}\right)+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}+\dfrac{1}{2}\cdot\dfrac{49}{100}-\dfrac{49}{99}\)
\(=\dfrac{49}{200}\)


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

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}=\frac{98}{99}\)
vậy \(A=\frac{98}{99}:2=\frac{49}{99}\)
chúc bạn học tốt
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

1/1x3 + 1/3x5 + 1/5x7 + ...............................+ 1/97x99
=1-1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.............................+ 1/97-1/99
=1-1/99
=98/99

\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)
\(\frac{99}{20}-2x=\frac{49}{99}\)
\(2x=\frac{99}{20}-\frac{49}{99}\)
\(2x=\frac{8821}{1980}\)
\(x=\frac{8821}{1980}:2\)
\(x=\frac{8821}{3960}\)

1/1.3 - 1/2.4 + 1/3.5 - 1/4.6 + ...+ 1/97.99 - 1/98.100 + 3lxl = 1
tách ra trc đầu tiên tính phần : 1/1.3 - 1/2.4 + 1/3.5 - 1/4.6 + ...+ 1/97.99 - 1/98.100
tách số lẻ và số chẵn ra
(1/1.3+1/3.5+...+1/57.97+1/97.99)-(1/2.4+1/4.6+...1/98.100)
tính từng vế vế đầu kết qu3 vế lẻ là : 49/99
kết quả vế chẵn là 49/200
thì bài đó sẻ thành : 49/99+49/200+3lxl=1
còn lại tự tinh nha
\(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{97\cdot99}\right)\cdot x=\frac{1}{99}\)
=>\(\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{97\cdot99}\right)\cdot x=\frac{2}{99}\)
=>\(\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{97}-\frac{1}{99}\right)\cdot x=\frac{2}{99}\)
=>\(x\cdot\frac{98}{99}=\frac{2}{99}\)
=>98x=2
=>\(x=\frac{2}{98}=\frac{1}{49}\)
(1/1.3+1/3.5+...+1/97.99).x=1/99
(1-1/3+1/3-1/5+...+1/97-1/99).x=1/99
(1-1/99).x=1/99
98/99.x=1/99
x=1/99 : 98/99
x=1/98