A=\(\frac{7}{10}\)*(\(\frac{10}{3\cdot13}\)+\(\frac{10}{13\cdot23}\)+\(\frac{10}{23\cdot33}\)+\(\frac{10}{43\cdot53}\)+\(\frac{10}{53\cdot63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{23}\)+\(\frac{1}{23}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{53}\)+\(\frac{1}{53}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{10}{33}\)+\(\frac{20}{2709}\))
A=\(\frac{7}{10}\)*\(\frac{9250}{29799}\)
A=\(\frac{925}{4257}\)
CM công thức :
\(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
\(A=\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}+...+\dfrac{7}{53.63}\)
\(A=\dfrac{7}{10}\left(\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}+...+\dfrac{7}{53.63}\right)\)
\(A=\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{33}+...+\dfrac{1}{53}-\dfrac{1}{63}\right)\)
\(A=\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{63}\right)\)
\(A=\dfrac{7}{10}.\dfrac{20}{63}=\dfrac{2}{9}\)
vậy A=\(\dfrac{2}{9}\)
chết quên !!!
cho mk sửa : ở dòng thứ 3 cho mk sửa tử trong ngoặc là 10 nha!!!!
XIN TRÂN THÀNH CẢM ƠN!!!