\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)

\(\frac{99}{20}-2x=\frac{49}{99}\)

\(2x=\frac{99}{20}-\frac{49}{99}\)

\(2x=\frac{8821}{1980}\)

\(x=\frac{8821}{1980}:2\)

\(x=\frac{8821}{3960}\)

\(=5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=5\left(1-\dfrac{1}{100}\right)\)

\(=5.\dfrac{99}{100}=\dfrac{99}{20}\)

A = 5/1.2 + 5/2.3 +...+ 5/99.100

 2A = 10/1.2 + 10/2.3 +...+ 10/99.100

2 A = 5/1-5/2+5/2-5/3+5/3-5/4+...+5/99-5/100

2A=5/1-5/100

2A=9/2 => A=9/2:2=9/4

cho 1 đ-ú-n-g nha bạn!!

99/20 đảm bảo

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Kẻ Bí Mật sai rồi !                                                                                      

Cai phan 1+3+5+...+99 chac em biet lam roi phai ko? Con 3/1.2+3/2.3+3/3.4+...+3/99.100 thi em cu tach lam sao cho tro thanh dang ban dau thi lam . Anh phai nghi roi !~ Neu chieu anh ranh ranh thi len giai tiep . BYE BYE

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}+4x=1+3+5+...+99\)

\(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+4x=\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\)\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+4x=100+100+100+...+100\)\(3\left(1-\frac{1}{100}\right)+4x=100.25\)

\(3.\frac{99}{100}+4x=2500\)

\(\frac{297}{100}+4x=2500\)

\(4x=2500-\frac{297}{100}\)

\(4x=2500-2,97\)

\(4x=2497,03\)

\(x=624,2575\)

\(x=2497,03:4\)

1/5 .B=1/1×2+1/2×3+.....+1/99×100

1/5 .B=2-1/1×2+3-2/2×3+......+100-99/99×100

1/5 .B=1/1-1/2+1/2-1/3+......+1/99-1/100

1/5.B=1-1/100

1/5. B=99/100

B=99/100÷1/5

B=99/20

\(B=\frac{5}{1.2}+\frac{5}{2.3}+.....+\frac{5}{99.100}\)

\(B=5-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+.....+\frac{5}{99}-\frac{5}{100}\)
\(B=5-\frac{5}{100}\)

\(B=\frac{500}{100}-\frac{5}{100}\)

\(B=\frac{495}{100}=\frac{99}{20}\)

1)C=5/1.2+5/2.3+5/3.4+...+5/99.100

   C=5.(1/1.2+1/2.3+1/3.4+...+1/99.100)

   C=5.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)

   C=5.(1/1-1/100)

   C=5.99/100

   C=99/20

2)|x+1|=5

⇒x+1=5 hoặc x+1=-5

       x=4 hoặc x=-6

  3)                    Giải:

Để A=2n+5/n+3 là số nguyên thì 2n+5 ⋮ n+3

2n+5 ⋮ n+3

⇒2n+6-1 ⋮ n+3

⇒1 ⋮ n+3

Ta có bảng:

n+3=-1 ➜n=-4

n+3=1 ➜n=-2

Vậy n ∈ {-4;-2}

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+.........+\frac{1}{99.100}\right)\)

\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{99}-\frac{1}{100}\right)\)

\(=5\left(1-\frac{1}{100}\right)\)

\(=5.\frac{99}{100}\)

\(=\frac{99}{20}\)

A =\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3\cdot4}+...+\frac{5}{99.100}\)

A = 5 x (\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) )

A = 5 x \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

A = 5 x \(\left(1-\frac{1}{100}\right)\)

A = 5 x \(\frac{99}{100}\)

A = \(\frac{495}{100}\)

A= \(\frac{99}{20}\)

Ta co : A =5.(1/1.2+1/2.3+1/3.4+....+1/99.100)

             A= 5.(1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100)

Rut gon tung so ta co :A=5.(1-1/100)

                                         A=5.99/100

                                          A=1.99/50=99/50

minh lon phai la20 o mau cua ket qua do