câu a: 14a + 6b = 84 + ab
<=> 14a + 6b - 84 - ab =0
<=> (14a -84) + (6b -ab)=0
<=> 14( a- 6) - b(a-6)=0
<=> (a - 6)(14-b) = 0
Vậy a=6, b=14
Đặt \(A=\dfrac{n}{4+n^4}\)
\(=\dfrac{n}{n^4+4n^2+4-4n^2}\)
\(=\dfrac{n}{\left(n^2+2\right)^2-\left(2n\right)^2}\)
\(=\dfrac{n}{\left(n^2+2-2n\right)\left(n^2+2+2n\right)}\)
\(\Rightarrow4A=\dfrac{4n}{\left(n^2-2n+2\right)\left(n^2+2n+2\right)}\)
\(=\dfrac{1}{n^2-2n+2}-\dfrac{1}{n^2+2n+2}\)
Đặt \(P=\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2n-1}{4+\left(2n-1\right)^4}\)
\(\Rightarrow4P=\dfrac{4}{4+1^4}+\dfrac{12}{4+3^4}+...+\dfrac{4\left(2n-1\right)}{4+\left(2n-1\right)^4}\)
\(=\dfrac{1}{1^2-2\times1+2}-\dfrac{1}{1^2+2\times1+2}\)
\(+\dfrac{1}{3^2-2\times3+2}-\dfrac{1}{3^2+2\times3+2}+...+\)
\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{17}+...+\)
\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{4n^2-4n+1+4n-2+2}\)
\(=1-\dfrac{1}{4n^2+1}\)
\(\Rightarrow P=\dfrac{1}{4}-\dfrac{1}{4\left(4n^2+1\right)}\)
