Câu này giống câu bên dưới bạn hỏi, bạn tham khảo cách mình làm ạ !

Em chỉ thử thôi, giáo viên nào đi qua check hộ em với ạ!

\(y'=3-\dfrac{4}{x^3}\)

\(y'=0\Leftrightarrow x=\sqrt[3]{\dfrac{4}{3}}\notin[2;+\infty)\)

\(\Rightarrow y_{min}=f\left(2\right)=3.6+\dfrac{2}{2^2}=\dfrac{13}{2}\)

b/ \(y'=2-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow x=1\notin(0;\dfrac{2}{3}]\)

\(f\left(0,4\right)=7,05;f\left(0,5\right)=5\Rightarrow ham-nghich-bien-trong-nua-khoang-(0;\dfrac{2}{3}]\)

\(\Rightarrow y_{min}=f\left(\dfrac{2}{3}\right)=2.\dfrac{2}{3}+\dfrac{1}{\left(\dfrac{2}{3}\right)^2}=\dfrac{43}{12}\)

c/ \(y=x+\dfrac{1}{x-1}\Rightarrow y'=1-\dfrac{1}{\left(x-1\right)^2}\)

\(y'=0\Leftrightarrow1-\dfrac{1}{\left(x-1\right)^2}=0\Leftrightarrow\left[{}\begin{matrix}x=0\notin\left(1;+\infty\right)\\x=2\in\left(1;+\infty\right)\end{matrix}\right.\)

\(f\left(\dfrac{3}{2}\right)=\dfrac{3}{2}+\dfrac{1}{\dfrac{3}{2}-1}=\dfrac{7}{2};f\left(2\right)=3;f\left(3\right)=\dfrac{7}{2}\)

=> ham nghich bien tren \(\left(1;2\right)\) va dong bien tren \([2;+\infty)\)

\(\Rightarrow y_{min}=f\left(2\right)=3\)

d/ \(y=\dfrac{1}{x}+\dfrac{2}{1-x}\Rightarrow y'=-\dfrac{1}{x^2}+\dfrac{2}{\left(1-x\right)^2}\)

\(y'=0\Leftrightarrow-\dfrac{1}{x^2}+\dfrac{2}{\left(1-x\right)^2}=0\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\in\left(0;1\right)\\x=-1-\sqrt{2}\notin\left(0;1\right)\end{matrix}\right.\)

\(f\left(0,2\right)=\dfrac{15}{2};f\left(\sqrt{2}-1\right)=3+2\sqrt{2};f\left(0,5\right)=6\)

=> f(x) nghich bien tren \(\left(0;\sqrt{2}-1\right)\)

dong bien tren \([\sqrt{2}-1;1)\)

\(\Rightarrow y_{min}=f\left(\sqrt{2}-1\right)=3+2\sqrt{2}\)

e/ \(y=\dfrac{x^2+2x+2}{x+1}\Rightarrow y'=\dfrac{\left(x^2+2x+2\right)'\left(x+1\right)-\left(x+1\right)'\left(x^2+2x+2\right)}{\left(x+1\right)^2}\)

\(y'=\dfrac{\left(x+1\right).\left(2x+2\right)-x^2-2x-2}{\left(x+1\right)^2}=\dfrac{2x^2+4x+2-x^2-2x-2}{\left(x+1\right)^2}=\dfrac{x^2+2x}{x^2+2x+1}\)

\(y'=0\Leftrightarrow x^2+2x=0\Leftrightarrow x=0\in\left(-1;+\infty\right)\)

\(f\left(-0,5\right)=\dfrac{5}{2};f\left(0\right)=2;f\left(1\right)=\dfrac{5}{2}\)

=> f(x) nghich bien tren \(\left(-1;0\right)\)

dong bien tren \([0;+\infty)\)

\(\Rightarrow y_{min}=f\left(0\right)=2\)

\(HSXD\Leftrightarrow\dfrac{8\cos2x-m}{\sin^2x-2\sin x+3}\ge0\)

Have: \(\sin^2x-2\sin x+3=\left(\sin x-1\right)^2+2>0\forall x\)

\(\Rightarrow8\cos2x-m\ge0\)

\(\Leftrightarrow m\le8\cos2x\Leftrightarrow m\le-8\)

=> có 52 giá trị của m t/m

\(\left|\overrightarrow{MA}+\overrightarrow{MB}\right|=2\left|\overrightarrow{MI}\right|\) nhé, đánh thiếu, nhưng nó ko ảnh hưởng gì đến bài toán :v

\(\overrightarrow{AB}\left(2;1\right);\overrightarrow{BC}\left(-3;1\right);\overrightarrow{CA}\left(1;-2\right)\)

\(ptts:\)

\(d_{AB}:\left\{{}\begin{matrix}x=2+2t\\y=t\end{matrix}\right.\)

\(d_{BC}:\left\{{}\begin{matrix}x=4-3t\\y=1+t\end{matrix}\right.\)

\(d_{CA}:\left\{{}\begin{matrix}x=1+t\\y=2-2t\end{matrix}\right.\)

\(pttq:\)

\(d_{AB}:-1\left(x-2\right)+2y=0\Leftrightarrow2y-x+2=0\)

\(d_{BC}:x-4+3\left(y-1\right)=0\Leftrightarrow x+3y-7=0\)

\(d_{CA}:2\left(x-1\right)+y-2=0\Leftrightarrow2x+y-4=0\)

b/ \(\overrightarrow{MB}=\overrightarrow{CM}\Rightarrow M\left(\dfrac{x_B+x_C}{2};\dfrac{y_B+y_C}{2}\right)\Rightarrow M\left(\dfrac{5}{2};\dfrac{3}{2}\right)\)

\(\Rightarrow\overrightarrow{AM}\left(\dfrac{1}{2};\dfrac{3}{2}\right)\Rightarrow\overrightarrow{n_{AM}}=\left(-\dfrac{3}{2};\dfrac{1}{2}\right)\)

\(\Rightarrow d_{AM}:-\dfrac{3}{2}\left(x-2\right)+\dfrac{1}{2}y=0\Leftrightarrow\dfrac{1}{2}y-\dfrac{3}{2}x+3=0\)

\(M\in\left(d_1\right)\Rightarrow M\left(x;\dfrac{x+3}{2}\right)\)

\(\left|\overrightarrow{MA}+\overrightarrow{MB}\right|=\left|\overrightarrow{MI}\right|\)      \(\left(\overrightarrow{IA}=\overrightarrow{BI}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=-\dfrac{1}{2}\\y_I=\dfrac{y_A+y_B}{2}=\dfrac{7}{2}\end{matrix}\right.\Rightarrow I\left(-\dfrac{1}{2};\dfrac{7}{2}\right)\)

\(\Rightarrow\left|\overrightarrow{MA}+\overrightarrow{MB}\right|_{min}\Leftrightarrow\left|\overrightarrow{MI}\right|_{min}\Leftrightarrow\overrightarrow{MI}\perp\overrightarrow{AB}\Leftrightarrow\overrightarrow{MI}.\overrightarrow{AB}=0\)

\(\Leftrightarrow\left(x_I-x_M;y_I-y_M\right).\left(x_B-x_A;y_B-y_A\right)=0\)

\(\Leftrightarrow\left(x_I-x_M\right)\left(x_B-x_A\right)+\left(y_I-y_M\right)\left(y_B-y_A\right)=0\)

\(\Leftrightarrow\left(-\dfrac{1}{2}-x\right).\left(-3\right)+\dfrac{7}{2}-\dfrac{x+3}{2}=0\Rightarrow M\left(...\right)\)

\(y=\dfrac{x^2+3x+3}{x^2+1}\Rightarrow y'=\dfrac{\left(x^2+3x+3\right)'\left(x^2+1\right)-\left(x^2+3x+3\right)\left(x^2+1\right)'}{\left(x^2+1\right)^2}\)

\(y'=\dfrac{\left(x^2+1\right)\left(2x+3\right)-\left(x^2+3x+3\right).2x}{\left(x^2+1\right)^2}\)

\(y'=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)-2x\left(x^2+3x+3\right)=0\)

\(\Leftrightarrow2x^3+3x^2+2x+3-2x^3-6x^2-6x=0\)

\(\Leftrightarrow3x^2+4x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=..\\x=...\end{matrix}\right.\)

Check lai ho t nhe

Bà nói vậy thì chắc có mỗi tui mặt dày xin được cái gmail của ảnh quá :D.

_{P/s:Tui hỏi thế tại tui sợ ảnh nghỉ làm ở hoc24 á bà :( Nâuuuu, thế thì puồn lắm.}

\(=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+\pi^{21}\right)\left(1-2x\right)^{\dfrac{1}{7}}-\pi^{21}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{7}\left(1-2x\right)^{-\dfrac{6}{7}}.\left(-2\right)\left(x^2+\pi^{21}\right)+2x\left(1-2x\right)^{\dfrac{1}{7}}}{1}\)

\(=\dfrac{1}{7}.\left(-2\right).\pi^{21}=...\)

À, Ck iu là một thằng ất ơ nào đó tưởng tui là con trai :v Explain rất nhiều mà nó éo tin, nên t quyết định chơi bê đê với nó luôn :v Nhỉ Ck nhỉ :3 . Bà có bà Vy Le rồi, đừng chơi bắt cá bằng lưới nha :D

T tưởng bà biết in4 anh Lâm là hỏi được chớ :D Tại nick hoc24 ảnh ko onl mà gmail cũng off 4 ngày r :((