Question

a , Phân tích \(a^4+4\)thành nhân tử

b, Hãy tính \(B=\dfrac{3^4+4}{5^4+4}\times\dfrac{7^4+4}{9^4+4}....\dfrac{19^4+4}{21^4+4}\)

Answer 1

a) Ta có: \(a^4+4=a^4+4a^2+4-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2=\left(a^2+2-2a\right)\left(a^2+2+2a\right)\)

Answer 2

ta thấy n⁴+4=(n²-2n+2)(n²+2n+2)=\(\left[\left(n-1\right)^2-1\right]\) \(\left[\left(n+1\right)^2+1\right]\)

Do đó B=\(\dfrac{\left(2^2+1\right)\left(4^2+1\right)}{\left(4^2+1\right)\left(6^2+1\right)}.\dfrac{\left(6^2+1\right)\left(8^2+1\right)}{\left(8^2+1\right)\left(10^2+1\right)}.....\dfrac{\left(18^2+1\right)\left(20^2+1\right)}{\left(20^2+1\right)\left(22^2+1\right)}=\dfrac{2^2+1}{22^2+1}=\dfrac{5}{485}=\dfrac{1}{97}\)