Question

a) Phân tích đa thức sau thành nhân tử

a⁴ +a

b)tính A=\(\frac{\left(1^4+4\right)\left(5^4+4\right).......\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)........\left(23^4+4\right)}\)

Answer 1

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

Answer 2

a.\(a^4+a=a\left(a^3+1\right)=a\left(a+1\right)\left(a^2-a+1\right)\)

Answer 3

cám ơn bn nha