Question

[maianh nguyễn]

Câu 1:

Xác đinh k để hàm: f(x)=\(\left\{{}\begin{matrix}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}\\k\end{matrix}\right.\)liên tục tại 1

Câu 2: Cho \(lim\)(x-->1) \(\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\). Tổng S= \(a^2+b^2\) bằng bao nhiêu

Câu 3: lim(x->1) \(\frac{\sqrt{x^2+x+2}-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\frac{a\sqrt{2}}{b}+c\) với a/b là phân số tối giản. Tính a+b+c

Answer 1

[Nguyễn Việt Lâm]

\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)

Để hàm liên tục tại \(x=1\)

\(\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow k=2\sqrt{2019}\)

2.

\(\lim\limits_{x\rightarrow1}\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}a+b+1=0\\\lim\limits_{x\rightarrow1}\frac{2x+a}{2x}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-1\\\frac{a+2}{2}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=0\end{matrix}\right.\) \(\Rightarrow S=1\)

3.

\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)

\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)

\(\Rightarrow a+b+c=1+12+0=13\)