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\(A=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\left(\sqrt[5]{1-5x}-1\right)+x^2}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{5x\left(x^2+2017\right)}{\sqrt[5]{\left(1-5x\right)^4}+\sqrt[5]{\left(1-5x\right)^3}+\sqrt[5]{\left(1-5x\right)^2}+\sqrt[5]{1-5x}+1}+x^2}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(-\dfrac{5\left(x^2+2017\right)}{\sqrt[5]{\left(1-5x\right)^4}+\sqrt[5]{\left(1-5x\right)^3}+\sqrt[5]{\left(1-5x\right)^2}+\sqrt[5]{1-5x}+1}+x\right)\)
\(=-2017\)
dễ thấy hàm số trên có dạng 0/0
áp dụng quy tắc l'Hôpital
\(A=_{\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\sqrt[5]{1-5x}-2017}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\left(x^2+2017\right)\sqrt[5]{1-5x}-2017\right)'}{\left(x\right)'}}\)
\(A=\lim\limits_{x\rightarrow0}\dfrac{-x^2-2017}{\sqrt[5]{\left(1-5x\right)^4}}+2x\sqrt[5]{1-5x}=\dfrac{-2017}{1}=-2017\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{-a\sqrt{1+\dfrac{1}{x^2}}+\dfrac{2017}{x}}{1+\dfrac{2018}{x}}=-a\Rightarrow a=-\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{bx+1}{\sqrt{x^2+bx+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{b+\dfrac{1}{x}}{\sqrt{1+\dfrac{b}{x}+\dfrac{1}{x^2}}+1}=\dfrac{b}{2}=2\Rightarrow b=4\)
\(\Rightarrow P=2\)
Xét khai triển:
\(\left(1+x\right)^{2017}=C_{2017}^0+xC_{2017}^1+x^2C_{2017}^2+...+x^{2017}C_{2017}^{2017}\)
Lấy tích phân 2 vế:
\(\int\limits^1_0\left(1+x\right)^{2017}=\int\limits^1_0\left(C_{2017}^0+xC_{2017}^1+...+x^{2017}C_{2017}^{2017}\right)\)
\(\Leftrightarrow\dfrac{2^{2018}-1}{2018}=C_{2017}^0+\dfrac{1}{2}C_{2017}^1+...+\dfrac{1}{2018}C_{2017}^{2017}\)
Vậy \(S=\dfrac{2^{2018}-1}{2018}\)
Giới hạn đã cho hữu hạn khi \(2x^2+ax+b=0\) có nghiệm \(x=1\)
\(\Rightarrow2+a+b=0\Rightarrow b=-a-2\)
Ta được: \(\lim\limits_{x\rightarrow1}\dfrac{2x^2+ax-a-2}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(x+1\right)+a\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x+2+a\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{2x+2+a}{x+1}\)
\(=\dfrac{4+a}{2}=\dfrac{1}{4}\)
\(\Rightarrow a=-\dfrac{7}{2}\Rightarrow b=\dfrac{3}{2}\)
