Question

[Pum Nhố ll xD Saint x]

Rút gọn phân thức P=\(\frac{\left(1^4+4\right)\left(5^4 +4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Answer 1

[Phương Mai]

\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)

\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)

Answer 2

[Phương Mai]

cái bước tách ra bn nhân lại là có kết quả y chang, VD:

\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)