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Ta có : \(n^4+4=\left[\left(n-1\right)^2+1\right]\left[\left(n+1\right)^2+1\right]\)
Do đó :
\(M=\frac{1\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\frac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.\frac{\left(8^2+1\right)\left(10^2+1\right)}{\left(10^2+1\right)\left(12^2+1\right)}...\frac{\left(16^2+1\right)\left(18^2+1\right)}{\left(18^2+1\right)\left(20^2+1\right)}\)
\(M=\frac{1}{20^2+1}=\frac{1}{401}\)
Sửa đề: \(M=\frac{1^4+4}{3^4+4}\cdot\frac{5^4+4}{7^4+4}\cdot\frac{9^4+4}{11^4+4}\cdot...\cdot\frac{17^4+4}{19^4+4}\)
=\(\frac{\left(1^4+4\right)\cdot\left(5^4+4\right)\cdot\left(9^4+4\right)\cdot...\cdot\left(17^4+4\right)}{\left(3^4+4\right)\cdot\left(7^4+4\right)\cdot\left(11^4+4\right)\cdot...\cdot\left(19^4+4\right)}\)
\(=\frac{1\cdot17\cdot13\cdot145\cdot257}{17\cdot65\cdot29\cdot257\cdot401}=1\cdot\frac{1}{5}\cdot5\cdot\frac{1}{401}=\frac{1}{401}\)
Vậy: \(M=\frac{1}{401}\)
Công thức tổng quát ''mở'' cho bài toán trên được hình thành trên cơ sở phân tích thành nhân tử và được phát biểu như sau:
\(a^4+4=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)
Khi đó, biểu thức \(A\) trở thành:
\(A=\frac{\left(1^2-2+2\right)\left(1^2+2+2\right)\left(5^2-2.5+2\right)\left(5^2+2.5+2\right)...\left(17^2-2.17+2\right)\left(17^2+2.17+2\right)}{\left(3^2-2.3+2\right)\left(3^2+2.3+2\right)\left(7^2-2.7+2\right)\left(7^2+2.7+2\right)...\left(19^2-2.19+2\right)\left(19^2+2.19+2\right)}\)
\(A=\frac{\left(1^2-2+2\right)}{\left(19^2+2.19+2\right)}=\frac{1}{401}\)
Ta có:
\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)
\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)
\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)
\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)
...
\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)
\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)
=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)
Ta có: \(n^4+4=n^4+4n^2+4-4n^2=\left(n^2+2\right)^2-\left(2n\right)^2=\left(n^2+2n+2\right)\left(n^2-2n+2\right)\)
\(=\left[\left(n+1\right)^2+1\right]\left[\left(n-1\right)^2+1\right]\)
\(B=\frac{1^4+4}{3^3+4}\times\frac{5^4+4}{7^4+4}\times\frac{9^4+4}{11^4+4}\times...\times\frac{17^4+4}{19^4+4}\)
\(=\frac{2^2+1}{\left(2^2+1\right)\left(4^2+1\right)}\times\frac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}\times\frac{\left(8^2+1\right)\left(10^2+1\right)}{\left(10^2+1\right)\left(12^2+1\right)}\times...\times\frac{\left(16^2+1\right)\left(18^2+1\right)}{\left(18^2+1\right)\left(20^2+1\right)}\)
\(=\frac{1}{20^2+1}=\frac{1}{401}\)