Đặt \(A=\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)....\left(1+\dfrac{7}{2900}\right)\)
\(B=\left(81-\dfrac{3}{4}\right)\left(81-\dfrac{3^2}{5}\right)\left(81-\dfrac{3^3}{6}\right)....\left(81-\dfrac{3^{2014}}{2017}\right)\)
Ta có:
\(A=\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right).....\left(1+\dfrac{7}{2900}\right)\)
\(A=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.....\dfrac{2907}{2900}\)
\(A=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.....\dfrac{51.57}{50.58}\)
\(A=\dfrac{2.3.4.5.6....56.57}{1.2.3.4.5.....57.58}=\dfrac{1}{58}\)
\(B=\left(81-\dfrac{3}{4}\right)\left(81-\dfrac{3^2}{5}\right).....\left(81-\dfrac{3^{2014}}{2017}\right)\)
Vì trong dãy số trên có một thừa số là \(\left(81-\dfrac{3^6}{9}\right)=\left(81-81\right)=0\)
\(\Rightarrow B=0\)
Vì \(a=A+B\Rightarrow a=\dfrac{1}{58}+0=\dfrac{1}{58}\)(1)
Thay (1) vào đa thức \(f\left(x\right)=5x-29a\) ta được:
\(f\left(x\right)=5x-29.\dfrac{1}{58}=5x-\dfrac{1}{2}\)
Ta lại có:
\(f\left(x\right)=0\Leftrightarrow5x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{10}\)
Vậy nghiệm của đa thức trên là \(\dfrac{1}{10}\)
Chúc bạn học tốt!!!
