a: \(y=\left(x-1\right)^3\)

=>\(y'=\left[\left(x-1\right)^3\right]'=3\left(x-1\right)^2\cdot\left(x-1\right)'\)

\(=3\left(x-1\right)^2\)

b: \(y=\left(x+2\right)\left(2x^2-3\right)\)

=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)

=>\(y'=2x^2-3+2\left(x+2\right)\)

\(=2x^2+2x+1\)

c: \(y=\left(x-1\right)^2\left(x+2\right)\)

=>\(y=\left(x^2-2x+1\right)\left(x+2\right)\)

=>\(y'=\left(x^2-2x+1\right)'\left(x+2\right)-\left(x^2-2x+1\right)\left(x+2\right)'\)

=>\(y'=\left(2x-2\right)\left(x+2\right)-x^2+2x-1\)

\(=2x^2+4x-2x-4-x^2+2x-1\)

=>\(y'=x^2+4x-5\)

c: \(y=\left(x^2-1\right)\left(2x+1\right)\)

=>\(y'=\left(x^2-1\right)'\left(2x+1\right)+\left(x^2-1\right)\left(2x+1\right)'\)

\(=2x\left(2x+1\right)+2\left(x^2-1\right)\)

\(=4x^2+2x+2x^2-2=6x^2+2x-2\)

a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

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

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

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

=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)

y'=0

=>\(\left(x+3\right)^2-4=0\)

=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)

=>(x+5)(x+1)=0

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

=>\(\left(x-2\right)^2-9=0\)

=>(x-2-3)(x-2+3)=0

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1

a: ĐKXĐ: \(\left(x+2\right)\left(x+3\right)>=0\)

=>\(\left[{}\begin{matrix}x>=-2\\x< =-3\end{matrix}\right.\)

\(y=\sqrt{\left(x+2\right)\left(x+3\right)}=\sqrt{x^2+5x+6}\)

=>\(y'=\dfrac{\left(x^2+5x+6\right)'}{2\sqrt{x^2+5x+6}}=\dfrac{2x+5}{2\sqrt{x^2+5x+6}}\)

y'>0

=>\(\dfrac{2x+5}{2\sqrt{x^2+5x+6}}>0\)

=>2x+5>0

=>\(x>-\dfrac{5}{2}\)

Kết hợp ĐKXĐ, ta được: x>=-2

Đặt y'<0

=>2x+5<0

=>2x<-5

=>\(x< -\dfrac{5}{2}\)

Kết hợp ĐKXĐ, ta được: x<=-3

Vậy: Hàm số đồng biến trên \([-2;+\infty)\) và nghịch biến trên \((-\infty;-3]\)

b: ĐKXĐ: \(\dfrac{2x+1}{x-3}>=0\)

=>\(\left[{}\begin{matrix}x>3\\x< =-\dfrac{1}{2}\end{matrix}\right.\)

\(y=\sqrt{\dfrac{2x+1}{x-3}}\)

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

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

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

\(=-\dfrac{\dfrac{7}{\left(x-3\right)^2}}{2\sqrt{\dfrac{2x+1}{x-3}}}< 0\forall x\) thỏa mãn ĐKXĐ, trừ x=-1/2 ra

=>Hàm số luôn đồng biến trên \(\left(3;+\infty\right);\left(-\infty;-\dfrac{1}{2}\right)\)

c:

ĐKXĐ: x>=-3

 \(y=\left(x+1\right)\sqrt{x+3}\)

=>\(y'=\left(x+1\right)'\cdot\sqrt{x+3}+\left(x+1\right)\cdot\sqrt{x+3}'\)

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

=>\(y'=\sqrt{x+3}+\dfrac{x+1}{2\sqrt{x+3}}\)

=>\(y'=\dfrac{2x+6+x+1}{2\sqrt{x+3}}=\dfrac{3x+7}{2\sqrt{x+3}}\)

Đặt y'>0

=>3x+7>0

=>x>-7/3

Kết hợp ĐKXĐ, ta được: x>-7/3

Đặt y'<0

3x+7<0

=>x<-7/3

Kết hợp ĐKXĐ, ta được: \(-3< x< -\dfrac{7}{3}\)

Vậy: Hàm số đồng biến trên \(\left(-\dfrac{7}{3};+\infty\right)\) và nghịch biến trên \(\left(-3;-\dfrac{7}{3}\right)\)

d: \(y=\dfrac{x-1}{x^2+1}\)(ĐKXĐ: \(x\in R\))

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

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

Đặt y'>0

=>\(-x^2+2x+1>0\)

=>\(1-\sqrt{2}< x< 1+\sqrt{2}\)

Đặt y'<0

=>\(-x^2+2x-1< 0\)

=>\(\left[{}\begin{matrix}x>1+\sqrt{2}\\x< 1-\sqrt{2}\end{matrix}\right.\)

Vậy: hàm số đồng biến trên khoảng \(\left(1-\sqrt{2};1+\sqrt{2}\right)\)

hàm số nghịch biến trên khoảng \(\left(1+\sqrt{2};+\infty\right);\left(-\infty;1-\sqrt{2}\right)\)

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

b/ \(y'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\sqrt{x^3+1}-x^2+2\right)'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\dfrac{3x^2}{\sqrt{x^3+1}}-2x\right)\)c/ 

\(y'=14\left(x^6+2x-3\right)^6\left(x^6+2x-3\right)'=14\left(x^6+2x-3\right)^6\left(6x^5+2\right)\)

d/ \(y=\left(x^3-1\right)^{-\dfrac{5}{2}}\Rightarrow y'=-\dfrac{5}{2}\left(x^3-1\right)^{-\dfrac{7}{2}}\left(x^3-1\right)'=-\dfrac{15x^2}{2\sqrt{\left(x^3-1\right)^7}}\)

...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)

a: \(y=\left(x+2\right)^2=x^2+4x+4\)

=>\(y'=2x+4\)

Đặt y'>0

=>2x+4>0

=>x>-2

Đặt y'<0

=>2x+4<0

=>x<-2

Vậy: Hàm số đồng biến trên \(\left(-2;+\infty\right)\) và nghịch biến trên \(\left(-\infty;-2\right)\)

b: \(y=\left(x^2-1\right)\left(x+2\right)\)

=>\(y'=\left(x^2-1\right)'\cdot\left(x+2\right)+\left(x^2-1\right)\left(x+2\right)'\)

\(=2x\left(x+2\right)+x^2-1=2x^2+4x+x^2-1=3x^2+4x-1\)

Đặt y'>0

=>\(3x^2+4x-1>0\)

=>\(\left[{}\begin{matrix}x>\dfrac{-2+\sqrt{7}}{3}\\x< \dfrac{-2-\sqrt{7}}{3}\end{matrix}\right.\)

Đặt y'<0

=>\(3x^2+4x-1< 0\)

=>\(\dfrac{-2-\sqrt{7}}{3}< x< \dfrac{-2+\sqrt{7}}{3}\)

Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-2-\sqrt{7}}{3}\right);\left(\dfrac{-2+\sqrt{7}}{3};+\infty\right)\)

Hàm số nghịch biến trên khoảng \(\left(\dfrac{-2-\sqrt{7}}{3};\dfrac{-2+\sqrt{7}}{3}\right)\)

c: \(y=\left(x+2\right)\left(2x^2-3\right)\)

=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)

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

\(=6x^2+8x-3\)

Đặt y'>0

=>\(6x^2+8x-3>0\)

=>\(\left[{}\begin{matrix}x>\dfrac{-4+\sqrt{34}}{6}\\x< \dfrac{-4-\sqrt{34}}{6}\end{matrix}\right.\)

Đặt y'<0

=>\(6x^2+8x-3< 0\)

=>\(\dfrac{-4-\sqrt{34}}{6}< x< \dfrac{-4+\sqrt{34}}{6}\)

Vậy: hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-4-\sqrt{34}}{6}\right);\left(\dfrac{-4+\sqrt{34}}{6};+\infty\right)\)

Hàm số nghịch biến trên khoảng \(\left(\dfrac{-4-\sqrt{34}}{6};\dfrac{-4+\sqrt{34}}{6}\right)\)

d: \(y=\left(x-1\right)^2\left(x+2\right)\)

\(=\left(x^2-2x+1\right)\left(x+2\right)\)

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

=>\(y=x^3-3x+2\)

=>\(y'=3x^2-3\)

Đặt y'>0

=>\(3x^2-3>0\)

=>\(x^2>1\)

=>\(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

Đặt y'<0

=>\(3x^2-3< 0\)

=>x^2<1

=>-1<x<1

Vậy: Hàm số đồng biến trên các khoảng \(\left(1;+\infty\right);\left(-\infty;-1\right)\)

Hàm số nghịch biến trên khoảng (-1;1)

a. Làm gọn 1 chút xíu:

\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)

\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)

b.

 \(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)

c.

\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)

a: \(y=\left(x^2-1\right)^2\)

=>\(y'=2\left(x^2-1\right)'\left(x^2-1\right)\)

\(=4x\left(x^2-1\right)\)

Đặt y'>0

=>\(x\left(x^2-1\right)>0\)

TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x^2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\)

=>\(x>1\)

TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< 0\)

Đặt y'<0

=>\(x\left(x^2-1\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x^2< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\-1< x< 1\end{matrix}\right.\)

=>0<x<1

TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\x^2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\)

=>x<-1

Vậy: Hàm số đồng biến trên các khoảng \(\left(1;+\infty\right);\left(-1;0\right)\)

Hàm số nghịch biến trên các khoảng (0;1) và \(\left(-\infty;-1\right)\)

b: \(y=\left(3x+4\right)^3\)

=>\(y'=3\left(3x+4\right)'\left(3x+4\right)^2\)

\(\Leftrightarrow y'=9\left(3x+4\right)^2>=0\forall x\)

=>Hàm số luôn đồng biến trên R

c: \(y=\left(x+3\right)^2\left(x-1\right)\)

=>\(y=\left(x^2+6x+9\right)\left(x-1\right)\)

=>\(y'=\left(x^2+6x+9\right)'\left(x-1\right)+\left(x^2+6x+9\right)\left(x-1\right)'\)

=>\(y'=\left(2x+6\right)\left(x-1\right)+x^2+6x+9\)

=>\(y'=2x^2-2x+6x-6+x^2+6x+9\)

=>\(y'=3x^2-2x+3\)

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

=>\(y'=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{8}{9}\right)\)

=>\(y'=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{8}{3}>=\dfrac{8}{3}>0\forall x\)

=>Hàm số luôn đồng biến trên R

d: \(y=\left(2x+2\right)\left(x^3-1\right)\)

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

\(=2\left(x^3-1\right)+3x^2\left(2x+2\right)\)

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

\(=8x^3+6x^2-2\)

Đặt y'>0

=>\(8x^3+6x^2-2>0\)

=>\(x>0,46\)

Đặt y'<0

=>\(8x^3+6x^2-2< 0\)

=>\(x< 0,46\)

Vậy: Hàm số đồng biến trên khoảng tầm \(\left(0,46;+\infty\right)\)

Hàm số nghịch biến trên khoảng tầm \(\left(-\infty;0,46\right)\)

ĐKXĐ:

a.

\(x^2-16>0\Rightarrow\left[{}\begin{matrix}x>4\\x< -4\end{matrix}\right.\)

b.

\(x^2-2x+1>0\Rightarrow\left(x-1\right)^2>0\Rightarrow x\ne1\)

c.

\(\left(2-x\right)\left(x+1\right)>0\Rightarrow-1< x< 2\)

d.

\(\left(x^2-1\right)\left(x+5\right)>0\Rightarrow\left[{}\begin{matrix}-5< x< -1\\x>1\end{matrix}\right.\)