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mình ko tích lại đâu

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Đặt \(B=\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)

\(\Leftrightarrow B=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)

\(\Leftrightarrow2B=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{13}\right)=1-\frac{3}{13}=\frac{10}{13}\)

\(\Leftrightarrow A=1+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}=1+\frac{10}{13}=\frac{23}{13}\)

ê bạn ơi tìm y sao lại có x

\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)

\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)

\(\frac{2}{3}\cdot y=\frac{10}{3}\)

\(y=\frac{10}{3}:\frac{2}{3}\)

y=5

dấu nhân bạn ạ

a: =1/5-1/7=2/35

c: =3/5-1/7=21/35-5/35=16/35

\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{1}{3}+\dfrac{1}{25}\)

\(=\dfrac{28}{75}\)

cám ơi bạn

cảm ơn bạn rất nhiều

`1/15+1/35+1/63+1/99+1/143`

`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`

`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`

`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`

`=1/2.(1/3-1/13)`

`=1/2 . 10/39`

`=5/39`

a) \(\dfrac{3}{35}+\dfrac{7}{20}+\dfrac{32}{35}+\dfrac{13}{20}\)

Áp dụng tính chất giao hoán, kết hợp của phép cộng, ta được:

\(\left(\dfrac{3}{35}+\dfrac{32}{35}\right)+\left(\dfrac{7}{20}+\dfrac{13}{20}\right)\)

\(=1+1\)

\(=2\)

b) \(\dfrac{18\cdot15\cdot56}{42\cdot27\cdot25}=\dfrac{2\cdot3\cdot3\cdot5\cdot3\cdot4\cdot2\cdot7}{2\cdot3\cdot2\cdot2\cdot3\cdot9\cdot5\cdot5}\)

\(=\dfrac{2\cdot2\cdot3\cdot3\cdot3\cdot4\cdot5\cdot7}{2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5\cdot9}=\dfrac{3\cdot2\cdot2\cdot7}{2\cdot5\cdot9}=\dfrac{42}{45}\)