\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}\right)=\frac{20}{41}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(1-\frac{1}{x+2}=\frac{40}{41}\)

\(\frac{1}{x+2}=1-\frac{40}{41}\)

\(\frac{1}{x+2}=\frac{1}{41}\)

\(x+2=41\)

\(x=41-2\)

\(x=39\)

Tìm x 

a) \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{x\times\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+1\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{40}{41}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{\left(x+2\right)}=\frac{40}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Rightarrow x+2=41\)

\(\Rightarrow x=41-2\)

\(\Rightarrow x=39\)

Vậy x = 39

Gửi kết bn với mh nhé!

ban ko tra loi ho minh cau nay a 

neu the minh ket ban kieu gi

\(\left[12\cdot15-x\right]\cdot\frac{1}{4}=120\cdot\frac{1}{4}\)

\(\Leftrightarrow\left[180-x\right]\cdot\frac{1}{4}=30\)

\(\Leftrightarrow180-x=30:\frac{1}{4}\)

\(\Leftrightarrow180-x=120\)

\(\Leftrightarrow x=60\)

tl moi cau

8/3

8/3

A ) x = 5/24

B ) x = 7/15

X x \(\frac{1}{3}\)+ X x \(\frac{2}{3}\)= 3

X x ( \(\frac{1}{3}\)+ \(\frac{2}{3}\)) = 3

X x 1 = 3

X      = 3 : 1

X       = 3

X x 1/3 + X x 2/3 = 3 

X x (1/3 + 2/3) = 3 

X x  1              = 3 

X = 3 : 1 

X = 3 

\(\frac{2}{5}:\frac{3}{7}x\frac{3}{7}:\frac{2}{5}+1999=\)\(\frac{2}{5}x\frac{7}{3}x\frac{3}{7}x\frac{5}{2}+1999=\)\(\left(\frac{2x7x3x5}{5x3x7x2}\right)+1999=\)\(1+1999=2000\)

\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)\)\(=\left(\frac{2}{2}-\frac{1}{2}\right)\)\(x\left(\frac{3}{3}-\frac{1}{3}\right)\)\(x\left(\frac{4}{4}-\frac{1}{4}\right)\)\(x\left(\frac{5}{5}-\frac{1}{5}\right)\)\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}=\)\(\frac{1x2x3x4}{2x3x4x5}=\frac{1x\left(2x3x4\right)}{\left(2x3x4\right)x5}=\frac{1}{5}\)\(.\)

2/5:3/7 nhân 3/7:2/5+1999

=2/5 nhân 7/3 nhân 3/7nhân 5/2+1999

=14/15 nhân 15/14+1999

=1+1999

=2000

(1-1/2) nhân (1-1/3) nhân (1-1/4) nhân (1-1/5)

=1/2 nhân 2/3 nhân 3/4 nhân 4/5

=24/120

=1/5

       Câu hỏi của Nguyễn Hữu Hưng       

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(x=305\)