a) Ta có: I là trung điểm AB

\(\Rightarrow\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=\dfrac{-1+3}{2}=1\\y_I=\dfrac{y_A+y_B}{2}=\dfrac{-2+2}{2}=0\end{matrix}\right.\)

\(\Rightarrow I\left(1;0\right)\)

b) Ta có: G là trọng tâm tam giác ABC

\(\Rightarrow\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{-1+3+4}{3}=2\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{-2+2+1}{3}=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow G\left(2;\dfrac{1}{3}\right)\)

a)  Gọi E là trung điểm AB \(\Rightarrow\) \(\overrightarrow{IA}+\overrightarrow{IB}=2\overrightarrow{IE}\)

 \(\overrightarrow{IA}+\overrightarrow{IB}+3\overrightarrow{IC}=\overrightarrow{0}\)

\(2\overrightarrow{IE}+3\overrightarrow{IC}=\overrightarrow{0}\)

b) \(\left|\overrightarrow{MA}+\overrightarrow{MB}+3\overrightarrow{MC}\right|\)

\(=\left|\overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{MI}+\overrightarrow{IB}+3\overrightarrow{MI}+3\overrightarrow{IC}\right|\)

\(=5MI\)

\(\left|\overrightarrow{MA}+\overrightarrow{MB}+3\overrightarrow{MC}\right|min\Leftrightarrow MImin\)

                                           \(\Leftrightarrow\) M là hình chiếu của I trên d

Ta có:

\(\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IM}\) (1)

Mặt khác: I là trung điểm AM

\(\Rightarrow\overrightarrow{AM}=2\overrightarrow{IM}\) (2)

Từ (1) và (2)

\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{AM}\)

\(M=d_1\cap d_2\)

\(\Leftrightarrow\)M là nghiệm của hệ phương trình: \(\left\{{}\begin{matrix}x+2y-11=0\\5x-3y-3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\) 

\(\Leftrightarrow M\left(3;4\right)\)

\(\Rightarrow\sqrt{x^2+y^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\)

\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2\left(x-1\right)}}{\sqrt{x-1}-\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{x}{1-\sqrt{x-1}}=\dfrac{1}{1}=1\)

\(\Rightarrow\left\{{}\begin{matrix}4a-2b+c=2\left(1\right)\\a-b+c=-5\left(2\right)\\a+b+c=-1\left(3\right)\end{matrix}\right.\)

\(\left(2\right)+\left(3\right)\Leftrightarrow a+c=-3\) \(\Rightarrow b=2\)

\(\Rightarrow4a+c=2+4=6\)

\(\Rightarrow\left\{{}\begin{matrix}a+c=-3\\4a+c=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3\\c=-6\end{matrix}\right.\)

\(lim\left(\sqrt{9n^2+10n}-an\right)=-\infty\)

\(\Leftrightarrow lim\dfrac{9n^2+10n-a^2n^2}{\sqrt{9n^2+10n}}=-\infty\)

\(\Leftrightarrow lim\dfrac{9-a^2+\dfrac{10}{n}}{\sqrt{\dfrac{9}{n^2}+\dfrac{10}{n^3}}}=-\infty\)

\(\Leftrightarrow\dfrac{9-a^2}{0}=-\infty\)

\(\Rightarrow a^2>9\)

\(\Leftrightarrow a>3\) \(\Rightarrow a\in\left[4;2023\right]\)

\(lim\left(\sqrt[3]{n^3-2n^2}-2n+1\right)\)

\(=lim\dfrac{n^3-2n^2-\left(2n-1\right)^3}{\left(\sqrt[3]{n^3-2n^2}\right)^2+\sqrt[3]{n^3-2n^2}\left(2n-1\right)+\left(2n-1\right)^2}\)

\(=lim\dfrac{n^3-2n^2-\left(8n^3-12n^2+6n-1\right)}{\sqrt[3]{n^6-4n^5+4n^4}+\sqrt[3]{n^3-2n^2}\left(2n-1\right)+4n^2-4n+1}\)

\(=lim\dfrac{-7n^3+10n^2-6n+1}{\sqrt[3]{n^6-4n^5+4n^4}+\sqrt[3]{n^3-2n^2}\left(2n-1\right)+4n^2-4n+1}\)

\(=lim\dfrac{-7+\dfrac{10}{n}-\dfrac{6}{n^2}+\dfrac{1}{n^3}}{\sqrt[3]{\dfrac{1}{n^3}-\dfrac{4}{n^4}+\dfrac{4}{n^5}}+\sqrt[3]{\dfrac{1}{n^3}-\dfrac{2}{n^4}}\left(2-\dfrac{1}{n}\right)+\dfrac{4}{n}-\dfrac{4}{n^2}+\dfrac{1}{n^3}}\)

\(=\dfrac{-7}{0}=-\infty\)

\(\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\) (\(Đk:x\ge0\))
Ta có: \(\sqrt{x}+1\ge1\Rightarrow\dfrac{5}{\sqrt{x}+1}\le5\)

\(\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\le1+5=6\)

\(Max\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=6\Leftrightarrow x=0\)