Rút gọn: (sử dụng phương pháp thêm bớt cùng 1 hạng tử)
\(\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)...\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)...\left(20^4+\dfrac{1}{4}\right)}\)
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Đặt \(A=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)...\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)...\left(20^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left[\left(1^4+\dfrac{1}{4}\right).2^4\right]\left[\left(3^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(19^4+\dfrac{1}{4}\right).2^4\right]}{\left[\left(2^4+\dfrac{1}{4}\right).2^4\right]\left[\left(4^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(20^4+\dfrac{1}{4}\right).2^4\right]}\)
\(=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)
Lưu ý: \(a^4+4=\left(a^4+4a^2+4\right)-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2\)
\(=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)
Áp dụng vào biểu thức A, ta có:
\(A=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)
\(=\dfrac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)...\left(38^2-38.2+2\right)\left(38^2+38.2+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)...\left(40^2-2.40+2\right)\left(40^2+2.40+2\right)}\)
\(=\dfrac{2.10.26..1370.1522}{10.26.50...1522.1682}=\dfrac{2}{1682}=\dfrac{1}{841}\)
Vậy \(A=\dfrac{1}{841}\)