a) phân tích đa thức \(n^4+\frac{1}{4}\)thành nhân tử
(gợi ý:thêm bớt hạng tử)
b) Áp dụng: Rút gọn:
\(S=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
(kết quả là:\(\frac{1}{?}\))
Phân tích đa thức thành nhân tử :
\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
Thầy tớ có cho gợi ý ạ :
Có dạng tổng quát của mỗi thừa số của tử số và mẫu số : \(n^4+\frac{1}{4}\)
Phân tích thành : \(n^4+\frac{1}{4}\)
\(=\left(n^2\right)^2+2n^2.\frac{1}{2}+\frac{1}{4}-n^2\)
\(=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)
P/s : Giải giúp tớ nhé :33
Câu hỏi: Rút gọn biểu thức A = \(\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(\left(2k\right)^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(\left(2k-1\right)^4+\frac{1}{4}\right)}\) (k thuộc N*)
Cho \(n^4+\frac{1}{4}=\left(\left(n-1\right)n+\frac{1}{2}\right)\left(\left(n+1\right)n+\frac{1}{2}\right)\)
Thu gọn phân thức:
\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(13^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(14^4+\frac{1}{4}\right)}\)
Tính
\(S=\frac{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right).\left(5^4+\frac{1}{4}\right).....\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right).\left(6^4+\frac{1}{4}\right)....\left(20^4+\frac{1}{4}\right)}\)
Tính:
\(S=\frac{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right).\left(5^4+\frac{1}{4}\right).....\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right).\left(6^4+\frac{1}{4}\right)....\left(20^4+\frac{1}{4}\right)}\)
Rút gọn
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right).....\left(15^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right).....\left(16^4+\frac{1}{4}\right)}\)
Rút gọn :
\(A=\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(51^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(52^4+\frac{1}{4}\right)}\)
tính giá trị của biểu thức
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)....\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)....\left(20^4+\frac{1}{4}\right)}\)
Tính A=\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
