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Đặt :
\(PHUC=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)..........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).........\left(12^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(1.2+\dfrac{1}{2}\right).........\left(12.13+\dfrac{1}{2}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(\Leftrightarrow PHUC=\dfrac{1}{313}\)
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).....\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2+2^2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right).......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak
Bài 2:
\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)
\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)
\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)
\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)
\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)
\(=\dfrac{7.2.10}{81-4}\)
\(=\dfrac{720}{77}\)
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 \(A=\dfrac{1}{841}\)
mk thấy cũng khá đơn giản mà Phạm Ngọc Diễm