\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{30}{93}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 93 - 3
=> 2x = 90
=> x = 90 : 2
=> x = 45
Vậy x = 45
Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\)
\(2A=\frac{1}{3}-\frac{1}{2x+3}\)
\(A=\frac{2x}{3\left(2x+3\right)}:2\)
\(\Rightarrow\frac{2x}{3\left(2x+3\right)}:2=\frac{15}{93}\)
\(\frac{2x}{3\left(2x+3\right)}=\frac{15}{93}.2=\frac{30}{93}=\frac{10}{31}\)
\(\frac{2x}{6x+9}=\frac{10}{31}\)
\(\Rightarrow2x.31=10.\left(6x+9\right)\)
\(\Rightarrow62x=60x+90\)
\(62x-60x=90\)
\(2x=90\)
\(x=45\)
Vậy x = 45
Ủng hộ mk nha !!! ^_^
