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\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
#)Giải :
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{y}-\frac{1}{y+2}=\frac{50}{101}\)
\(1-\frac{1}{y+2}=\frac{50}{101}\)
\(\Leftrightarrow\frac{1}{y+2}=\frac{51}{101}\)
\(\Leftrightarrow y+2=\frac{101}{51}\)
\(\Leftrightarrow x=-\frac{1}{51}\)
Có: \(x=\dfrac{1}{9}+\dfrac{8}{116}=\dfrac{1}{9}+\dfrac{2}{29}=\dfrac{47}{261}\)
\(x-\left(\dfrac{2+2+2+2}{3+15+35+63}\right)=\dfrac{1}{9}\)
\(\Leftrightarrow x=\dfrac{1}{9}+\dfrac{2}{29}=\dfrac{47}{261}\)
\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\frac{10}{11}.y=\frac{4}{3}\)
\(\Rightarrow y=\frac{22}{15}\)
\(\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right).Y=4\)
\(\Rightarrow\frac{2}{9}.Y=4\)
\(\Rightarrow Y=4:\frac{2}{9}\)
\(\Rightarrow Y=18\)
Vậy số cần tìm là 18.
~Study well~
\(\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right).y=4\)
\(\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right).y=4\)
\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right).y=4\)
\(\left(\frac{1}{3}-\frac{1}{9}\right).y=4\)
\(\frac{2}{9}.y=4\)
\(y=18\)
Vậy \(y=18\)
=> 2/1x3 +2/3x5+2/5x7+2/7x9+...+2/nx(n+2)
=>1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9+...+1/n-1/n+2
=>1-1/n+2=100/101
1/n+2=1-100/101
1/n+2=1/101
=>n+2=101
=>n=101-2
=>n=99
( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
\(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)
\(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)
\(x\) = 4
cong trong ngoac truoc di roi lay 1 chia cho so trong ngoac la ra
tk minh nha, chuc ban hoc tot
\(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{8}{3}\)
hay \(x=\dfrac{4}{9}\cdot\dfrac{3}{8}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)
y x ( \(\dfrac{2}{3}\) + \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\)+ \(\dfrac{2}{63}\)) = 1
y x ( \(\dfrac{4}{5}\) + \(\dfrac{4}{45}\)) = 1
y x \(\dfrac{8}{9}\) = 1
y = 1 : 8/9
y = 9/8