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\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).....\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2+2^2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right).......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
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 ...
Đặt :
\(PHUC=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)..........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).........\left(12^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(1.2+\dfrac{1}{2}\right).........\left(12.13+\dfrac{1}{2}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(\Leftrightarrow PHUC=\dfrac{1}{313}\)
đặt biểu thức đã cho là A
Ta có : \(a^4+\dfrac{1}{4}\) \(=a^4+a^2+\dfrac{1}{4}-a^2\)
\(=\left(a^2+\dfrac{1}{2}\right)^2-a^2\)
\(=\left(a^2+a+\dfrac{1}{2}\right)\left(a^2-a+\dfrac{1}{2}\right)\)
Thay vào biểu thức đã cho ta được:
\(\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right)\left(3^2+3+\dfrac{1}{2}\right)\left(3^2-3+\dfrac{1}{2}\right)...\left(29^2+29+\dfrac{1}{2}\right)\left(29^2-29+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)\left(4^2+4+\dfrac{1}{2}\right)\left(4^2-4+\dfrac{1}{2}\right)...\left(30^2+30+\dfrac{1}{2}\right)\left(30^2-30+\dfrac{1}{2}\right)}\)
Lại có :
\(\left(k+1\right)^2-\left(k+1\right)+\dfrac{1}{2}\) \(=k^2+2k+1-k-1+\dfrac{1}{2}\)
\(=k^2+k+\dfrac{1}{2}\)
\(\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right)\left(3^2+3+\dfrac{1}{2}\right)\left(2^2+2+\dfrac{1}{2}\right)...\left(29^2+29+\dfrac{1}{2}\right)\left(28^2+28+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(1^2+1+\dfrac{1}{2}\right)\left(4^2+4+\dfrac{1}{2}\right)\left(3^2+3+\dfrac{1}{2}\right)...\left(30^2+30+\dfrac{1}{2}\right)\left(29^2+29+\dfrac{1}{2}\right)}\)
= \(\dfrac{1^2-1+\dfrac{1}{2}}{30^2+30+\dfrac{1}{2}}\)
= \(\dfrac{\dfrac{1}{2}}{30^2+30+\dfrac{1}{2}}\)
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).......\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right)........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
Ta có một số phân tích sau : \(a^4\)\(+\)\(4\)\(=\)\(\left(a^2-2a+2\right)\)\(\left(a^2+2a+2\right)\)
Nhân mỗi biểu thức trong ngoặc ở cả tử thức với \(16\)\(=\)\(2^4\), ta được :
\(A\)\(=\)\(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)
\(A\)\(=\)\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)...\left(58^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)\left(12^4+4\right)...\left(60^4+4\right)}\)
Kết hợp với phân tích nêu trên, khi đó :
\(A\)\(=\)\(\frac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)\left(6^2-2.6+2\right)\left(6^2+2.6+2\right)....\left(58^2-2.58+2\right)\left(58^2+2.58+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)\left(8^2-2.8+2\right)\left(8^2+2.8+2\right)....\left(60^2-2.60+2\right)\left(60^2+2.60+2\right)}\)
\(\Rightarrow\)\(A\)\(=\)\(\frac{2.10.26.50.82.122....3250.3482}{10.26.50.82.122....3482.3722}\)\(=\)\(\frac{2}{3722}\)\(=\)\(\frac{1}{1861}\)