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 \(P=\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2n-1}{4+\left(2n-1\right)^4}\)

\(\Rightarrow4P=\dfrac{4}{4+1^4}+\dfrac{12}{4+3^4}+...+\dfrac{4\left(2n-1\right)}{4+\left(2n-1\right)^4}\)

\(=\dfrac{1}{1^2-2\times1+2}-\dfrac{1}{1^2+2\times1+2}\)

\(+\dfrac{1}{3^2-2\times3+2}-\dfrac{1}{3^2+2\times3+2}+...+\)

\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{17}+...+\)

\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{4n^2-4n+1+4n-2+2}\)

\(=1-\dfrac{1}{4n^2+1}\)

\(\Rightarrow P=\dfrac{1}{4}-\dfrac{1}{4\left(4n^2+1\right)}\)

Chọn B

\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)

\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)

\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)

\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)

\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)

Lời giải:

Ta có: \(4+(2n-1)^4=[(2n-1)^2+2]^2-[2(2n-1)]^2\)

\(=[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]\)

\(\Rightarrow \frac{2n-1}{4+(2n-1)^4}=\frac{2n-1}{[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]}\)

\(=\frac{1}{4}\left(\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)}\right)\)

Do đó:

\(\frac{1}{4+1^4}=\frac{1}{4}(1-\frac{1}{5})\)

\(\frac{3}{4+3^4}=\frac{1}{4}(\frac{1}{5}-\frac{1}{17})\)

\(\frac{5}{4+5^4}=\frac{1}{4}(\frac{1}{17}-\frac{1}{37})\)

......

Do đó:

\(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{4+(2n-1)^4}=\frac{1}{4}(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{17}+...+\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)})\)

\(=\frac{1}{4}(1-\frac{1}{(2n-1)^2+2+2(2n-1)})=\frac{1}{4}(1-\frac{1}{(2n-1+1)^2+1})\)

\(=\frac{1}{4}(1-\frac{1}{4n^2+1})=\frac{n^2}{4n^2+1}\)

Ta có đpcm.

n=1 ; \(\dfrac{1}{4+1^4}=\dfrac{1}{5}=\dfrac{1^2}{4.^2+1}=\dfrac{1}{5};dung\)

giả sử n =k đúng \(\Leftrightarrow S=\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}=\dfrac{k^2}{4k^2+1}\) (*)

cần c/m đúng n =k+1 ;

c/m

với n=k+1

\(S=\left(\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}\right)+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)

từ (*) =>\(S=\dfrac{k^2}{4k^2+1}+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)

\(k+1=t\Leftrightarrow k=t-1\)

\(S=\dfrac{t^2-2t+1}{4\left(t^2-2t+1\right)+1}+\dfrac{2t-1}{4+\left(2t-1\right)^4}\)

\(S=\dfrac{t^2-2t+2}{4t^2-8t+5}+\dfrac{2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{\left(t^2-2t+1\right)\left(4t^2+1\right)+2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}\)\(S=\dfrac{t^2\left(4t^2-8t+5\right)}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{t^2}{\left(4t^2+1\right)}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\)

Vậy tổng trên đúng với k +1

theo Quy nạp ta có dpcm

\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)

@Luân Đào