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Ta có:
\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)
\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)
\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)
\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)
...
\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)
\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)
=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)
b.\(A=\)viết lại đề nha bn
\(A=\frac{1^4+4}{3^4+4}.\frac{5^4+4}{7^4+4}...\frac{21^4+4}{23^4+4}\)
\(A=\frac{\left(4.1-3\right)^4+4}{\left(4.1-1\right)^4+4}.\frac{\left(4.2-3\right)^4+4}{\left(4.2-1\right)^4+4}...\frac{\left(4.6-3\right)^4+4}{\left(4.6-1\right)^4+4}\)
\(A=\frac{16.1^2-32.1+17}{16.1^2+1}.\frac{16.2^2-32.2+17}{16.2^2+1}....\frac{16.6^2-32.6+17}{16.6^2+1}\)
\(A=\frac{1}{17}.\frac{17}{65}.\frac{65}{145}....\frac{401}{577}=\frac{1}{577}\)
tíck mình nha bn thanks
a.\(a^4+a=a\left(a^3+1\right)=a\left(a+1\right)\left(a^2-a+1\right)\)
\(A=\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).....\left(51^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)....\left(52^4+\frac{1}{4}\right)}\)
\(=\frac{\left(1+1+\frac{1}{2}\right)\left(1-1+\frac{1}{2}\right)....\left(11^2-11+\frac{1}{2}\right)}{\left(2+2^2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)....\left(12^2-12+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)....\left(11.12+\frac{1}{2}\right)}{\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(12.13+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{12.13+\frac{1}{2}}\)
\(=\frac{1}{313}\)
Chúc bạn học tốt !!!
