1/ \(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x-2\right)}{x+1}=-3\)

De ham so lien tuc tren R \(\Leftrightarrow\lim\limits_{x\rightarrow-1}f\left(x\right)=f\left(-1\right)\Leftrightarrow a=-3\)

2/ \(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)\left(x-2\right)}{x-2}=3\)

De ham so lien tuc tren R \(\Leftrightarrow\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\Leftrightarrow m=3\)

3/ \(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{-\left(x-2\right)\left(x-1\right)}{x-1}=1\)

De ham so lien tuc tren R \(\Leftrightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Leftrightarrow3+m=1\Leftrightarrow m=-2\)

Mình có một thắc mắc là tại sao mấy câu hỏi của bạn đều bị xóa đi, điển hình là 2 câu hình học hôm qua mình mới giải xong, hôm nay liền ko thấy đâu?

1/ \(\lim\limits_{x\rightarrow-2}f\left(x\right)=\lim\limits_{x\rightarrow-2}\dfrac{\left(x-2\right)\left(x+2\right)}{x+2}=-4\)

\(f\left(4\right)=\lim\limits_{x\rightarrow-2}f\left(x\right)=-4\) => Hàm số liên tục tại x=-2

2/ \(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{x+3}{x-1}=-1\)

\(f\left(-1\right)=2\ne\lim\limits_{x\rightarrow-1}f\left(x\right)\Rightarrow\) hàm số gián đoạn tại x=-1

\(\text{Lim}\dfrac{3^n-4^n+1}{2.4^n+2^n}=\lim\dfrac{\left(\dfrac{3}{4}\right)^n-1+\left(\dfrac{1}{4}\right)^n}{2+\left(\dfrac{2}{4}\right)^n}=-\dfrac{1}{2}\)

Kẻ MK vuông góc AC

\(\left\{{}\begin{matrix}MK\perp AC\subset\left(SAC\right)\\MK\perp SA\subset\left(SAC\right)\end{matrix}\right.\Rightarrow MK\perp\left(SAC\right)\)

\(\Rightarrow d\left(M,\left(SAC\right)\right)=KM=\dfrac{1}{2}AB=\dfrac{1}{2}\sqrt{16a^2-4a^2}=a\sqrt{3}\)

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

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

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

4/ \(y'=\left(\cos5x\right)'.\cos7x+\cos5x.\left(\cos7x\right)'=-5\sin5x.\cos7x-7\cos5x\sin7x\)

5/ \(y'=\left(\cos x\right)'\sin^2x+\cos x\left(\sin^2x\right)'=-\sin^3x+2\sin x.\cos^2x\)

6/ \(y'=\left(\tan^42x\right)'=4.\tan^32x.\dfrac{2}{\cos^22x}\)

7/ \(y'=\dfrac{2\sin x+2\cos x-2x.\cos x+2x\sin x}{\left(\sin x+\cos x\right)^2}\)

Ờm, bạn tự rút gọn nhé :) Mình đang hơi lười :b

\(h\left(x\right)=\dfrac{1}{2}\cos\left(\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\right)+\dfrac{1}{2}\)

\(\Rightarrow h'\left(x\right)=\dfrac{1}{2}\left[-\sin\left(\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\right)\right].\left(\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\right)'=-\dfrac{1}{2}.\dfrac{\left(2\sqrt{x}+4\right)'\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+4\right)\left(\sqrt{x}-3\right)'}{\left(\sqrt{x}-3\right)^2}.\sin\left(\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\right)\)

\(=-\dfrac{1}{2}.\dfrac{\dfrac{1}{\sqrt{x}}\left(\sqrt{x}-3\right)-\dfrac{1}{2\sqrt{x}}\left(2\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)^2}\sin\left(\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\right)=-\dfrac{1}{2}.\dfrac{-\dfrac{3}{\sqrt{x}}-\dfrac{2}{\sqrt{x}}}{\left(\sqrt{x}-3\right)^2}\sin\left(\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\right)\)

\(f'\left(x\right)=3x^2-4x\)

\(f'\left(x\right)>0\Leftrightarrow3x^2-4x>0\Rightarrow\left[{}\begin{matrix}x>\dfrac{4}{3}\\x< 0\end{matrix}\right.\)

\(f'\left(2\right)=4\) ; \(f\left(2\right)=0\)

Phương trình tiếp tuyến:

\(y=4\left(x-2\right)+0\Leftrightarrow y=4x-8\)