Toán lớp 3?Sao giống lớp 4,5 hơn!

Ta có:

\(A=\frac{5}{15}+...+\frac{5}{399}\)

\(\Rightarrow A=5.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(\Rightarrow A=\frac{5}{7}\)

\(A=\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+...+\frac{5}{399}\)

\(A=\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+\frac{5}{7\cdot9}+...+\frac{5}{19\cdot21}\)

\(A=\frac{5}{3\cdot5\cdot2}+\frac{5}{5\cdot7\cdot2}+\frac{5}{7\cdot9\cdot2}+...+\frac{5}{19\cdot21\cdot2}\)

\(A=\frac{5}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\right)\)

\(A=\frac{5}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(A=\frac{5}{2}\cdot\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(A=\frac{5}{2}\cdot\left(\frac{7}{21}-\frac{1}{21}\right)\)

\(A=\frac{5}{2}\cdot\frac{6}{21}\)

\(A=\frac{30}{42}=\frac{5}{7}\)

\(\frac{5}{3}+\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+\frac{5}{143}\)

\(=\frac{5}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{11\cdot13}\right)\)

\(=\frac{5}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{5}{2}\cdot\left(1-\frac{1}{13}\right)\)

\(=\frac{5}{2}\cdot\frac{12}{13}\)

\(=\frac{30}{13}\)

\(\frac{5}{3}+\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+\frac{5}{143}\)

\(=5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)

\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{13}\right)\)

\(=\frac{5}{2}.\frac{12}{13}\)

\(=\frac{30}{13}\)

Ta có:

    A=5/15+5/35+5/63+5/99+...+5/2915

=>A=5/3.5+5/5.7+5/7.9+5/9.11+...+5/53.55

=>A=5/2.(2/3.5+2/5.7+2/7.9+2/9.11+...+2/53.55)

=>A=5/2.(2/3-2/5+2/5-2/7+2/7-2/9+2/9-2/11+...+2/53-2/55)

=>A=5/2.(2/3-2/55)

=>A=5/2.104/165

=>A=52/33

Vậy A=52/33

OK!

thank

Mk ko chép lại đầu bài đâu,thông cảm nha mk chỉ biết giải ý B

3B=1.2.3+2.3.3+3.4.3+...+1000.1001.3

    =1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+1000.1001.(1002-999)

    =1.2.3-1.2.0+2.3.4-2.3.1+3.4.5-3.4.2+...+1000.1001.1002-1000.1001.999

    =(1.2.3+2.3.4+3.4.5+...+1000.1001.1002) - (1.2.0+2.3.1+3.4.2+...+1000.1001.999)

    =1000.1001.1002

=>B=(1000.1001.1002):3

      =334 334 000

k hộ mk nha!

\(y=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

mình thắc mắc quy luật của phép tính trên là gì : 15 -> 35 -> 63 ... -> 399 ?

Quy luật la cac so o duoi mau cach nhau 2 don vi ( cach nhau mot so bang tu):

2/15=2/3.5 ( 3 va 5 cach nhau 2don vi)

2/35=2/5.7 ( 5 va 7 cach nhau 2 don vi)

.........

(Cai nay minh gui cho ban Duy Thanh nhe!!!!!!!!!!!!!!!!!!!)

\(\frac{2}{15}+\frac{2}{35}+...+\frac{2}{399}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{2}{7}\)

HT

     D =3-3²-....-3¹⁰⁰

(=)D=3-(3²+3³+....+3¹⁰⁰)

(=)3D=3.3-(3³+....+3¹⁰¹)

(=)2D=6-(3¹⁰¹-3²)

(=)D=6-(3¹⁰¹-3²) :2

     B=-2/15-2/35-....-2/399

(=)B=-2/15-(2/35-...-2/399)

(=)B=-2/15-(2/5.7-...-2/19.21)

(=)B=-2/15-(1/5-1/21)

(=)B=-2/15-16/105

(=)B=-2/7

giúp mình ik nhá

\(A=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)

\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)

\(=5\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{81-80}{80.81}\right)\)

\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)

\(=5\left(1-\frac{1}{81}\right)=\frac{5.80}{81}=\frac{400}{81}\)

b)

\(B=7\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\right)\)

\(=7.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{21.23}\right)\)

=> \(2.B=7\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)

\(=7\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{23-21}{21.23}\right)\)

\(=7.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{21}-\frac{1}{23}\right)\)

\(=7\left(\frac{1}{3}-\frac{1}{23}\right)=\frac{7.20}{69}=\frac{140}{69}\)

=> \(B=\frac{140}{69}:2=\frac{70}{69}\)

a) SỬA LẠI ĐỀ : A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+...+\frac{5}{6480}\)

       = \(5.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)

       =\(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)

        = \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)

        = \(5.\left(1-\frac{1}{81}\right)\)

        = \(5.\frac{80}{81}\)

        = \(\frac{400}{81}\)

b) B = \(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+...+\frac{7}{483}\)

        = \(\frac{7}{3.5}+\frac{7}{5.7}+\frac{7}{7.9}+...+\frac{7}{21.23}\)

        = \(\frac{7}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)

        = \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{23}\right)\)

        = \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{23}\right)\)

        = \(\frac{7}{2}.\frac{20}{69}\)

        = \(\frac{70}{69}\)

\(\frac{35}{51}\)