#)Giải :
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{y}-\frac{1}{y+2}=\frac{50}{101}\)
\(1-\frac{1}{y+2}=\frac{50}{101}\)
\(\Leftrightarrow\frac{1}{y+2}=\frac{51}{101}\)
\(\Leftrightarrow y+2=\frac{101}{51}\)
\(\Leftrightarrow x=-\frac{1}{51}\)
#)Mình viết nhầm chỗ cuối nhé :P
là y chứ k ph x đâu
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
=> \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{y}-\frac{1}{y+2}=\frac{50}{101}\)
=> \(1-\frac{1}{y+2}=\frac{50}{101}\)
=> \(\frac{1}{y+2}=1-\frac{50}{101}\)
=> \(\frac{1}{y+2}=\frac{61}{101}\)
Áp dụng tính chất phân số bằng nhau ta có :
=> \(1.101=61.\left(y+2\right)\)
=> \(101=61.\left(y+2\right)\)
=> \(\left(y+2\right)=101:61\)
=> \(y+2=\frac{101}{61}\)
=> \(y=\frac{101}{61}-2\)
=> \(y=\frac{-21}{61}\)
Đặt A = \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(\Rightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{y}-\frac{1}{y+2}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(1-\frac{1}{y+2}\right)=\frac{1}{2}\cdot\frac{y+1}{y+2}=\frac{y+1}{2y+4}\)
\(\Rightarrow\frac{y+1}{2y+4}=\frac{50}{101}\)
\(\Rightarrow101\left(y+1\right)=50\left(2y+4\right)\)
\(\Rightarrow101y+101=100y+200\)
\(\Rightarrow y=200-101=99\)
Vậy y = 99
