Câu \(\mathbf{1}\): \(\\\) \(y=x^3+6x^2-15x+2035\) \(\\\) TXĐ: \(D=\mathbb{R}\) \(\\\) \(y'=3x^2+12x-15\) \(\\\) Xét \(y'=0\) \(\\\) \(\Rightarrow 3x^2+12x-15=0 \Leftrightarrow \left[ \begin{aligned} x&=1 \\ x&=-5 \end{aligned} \right.\) \(\\\) BBT: \(\\\) \(\begin{array}{|c|ccccccc|}\hline x&-\infty &&-5&&1&&+\infty \\ \hline y'&&+&0&-&0&+&\\ \hline &&&2085&&&&+\infty \\ y&&\nearrow &&\searrow &&\nearrow &\\ &-\infty &&&&2028&&\\ \hline \end{array}\) \(\\\) \(\Rightarrow \) Hàm số đạt cực tiểu tại \(x=1\) \(\\\) \(\Rightarrow y_{ct}=y(1)=1^3+6.1^2-15.1+2035=2028\) \(\\\) Câu \(\mathbf{2}\): Ta có:\(\\\) \(y=x^3-3mx^2+3(3m+4)x+2025\) \(\\\) TXĐ: \(D=\mathbb{R}\) \(\\\) \(y'=3x^2-6mx+3(3m+4)=3[x^2-2mx+(3m+4)]\) \(\\\) Hàm số đồng biến trên \((2;+\infty)\) khi \( y' \ge 0 \forall x \in (2;+\infty)\) \(\\\) \(\Leftrightarrow x^2-2mx+3m+4 \ge 0 \forall x \in (2;+\infty)\) \(\\\) \(\Leftrightarrow x^2+4 \ge m(2x-3) \forall x \in (2;+\infty)\) \(\\\) Vì \(x>2 \Rightarrow 2x-3>0\), cô lập \(m\): \(\\\) \(\Rightarrow m \le \dfrac{x^2+4}{2x-3} \forall x \in (2;+\infty)\) \(\\\) Xét \(g(x)=\dfrac{x^2+4}{2x-3}\) trên \((2;+\infty)\): \(\\\) \(g'(x)=\dfrac{2x(2x-3)-2(x^2+4)}{(2x-3)^2}=\dfrac{2x^2-6x-8}{(2x-3)^2}\) \(\\\) Xét \(g'(x)=0\) \(\\\) \(\Leftrightarrow 2x^2-6x-8=0 \Leftrightarrow \left[ \begin{aligned} x&=-1\ \notin (2;+\infty) (l) \\ x&=4\ (tm) \end{aligned} \right.\) \(\\\) BBT: \(\\\) \(\begin{array}{|c|ccccc|}\hline x&2&&4&&+\infty \\ \hline g^{\prime }(x)&&-&0&+&\\ \hline &8&&&&+\infty \\ g(x)&&\searrow &&\nearrow &\\ &&&4&&\\ \hline \end{array}\) \(\\\) \(\Rightarrow \min_{(2;+\infty)} g(x)=g(4)=4\) \(\\\) Để \(m \le g(x)\ \forall x \in (2;+\infty)\) thì \( m \le \min_{(2;+\infty)} g(x) \Rightarrow m \le 4\) \(\\\) Do \(m \in \mathbb{Z}^+ \Rightarrow m \in \{1;2;3;4\}\) \(\\\) \(\Rightarrow \sum m = 1+2+3+4=10\)