Rút gọn: A=\(\frac{\left(1^4+4\right).\left(5^4+4\right).\left(9^4+4\right).....\left(21^4+4\right)}{\left(3^4+4\right).\left(7^4+4\right).\left(11^4+4\right).....\left(23^4+4\right)}\)
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\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)
\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)
cái bước tách ra bn nhân lại là có kết quả y chang, VD:
\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)
\(A=\frac{\left(1^4+4\right).\left(5^4+4\right).\left(9^4+4\right).....\left(21^4+4\right)}{\left(3^4+4\right).\left(7^4+4\right).\left(11^4+4\right).....\left(23^4+4\right)}\)
\(\Leftrightarrow A=\frac{\left(1+4\right).\left(4^2+1\right).\left(6^2+1\right).\left(8^2+1\right).\left(10^2+1\right)....\left(20^2+1\right).\left(22^2+1\right)}{\left(2^2+1\right).\left(4^2+1\right).\left(6^2+1\right).\left(8^2+1\right).\left(10^2+1\right).\left(12^2+1\right)....\left(24^2+1\right)}\)
\(\Leftrightarrow A=\frac{1+4}{\left(2^2+1\right).\left(24^2+1\right)}=\frac{5}{5.\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)
Sai bạn nhé bạn phải cm là n^4+4=\(\left(\left(n-1\right)^2+1\right).\left(\left(n+1\right)^2+1\right)\))