Với hai số thực bất kì \(4< a< b\) ta có:

\(y\left(b\right)-y\left(a\right)=\sqrt{b-4}-\sqrt{a-4}+\sqrt{b+1}-\sqrt{a+1}\)

\(=\frac{b-a}{\sqrt{b-4}+\sqrt{a-4}}+\frac{b-a}{\sqrt{b+1}+\sqrt{a+1}}=\left(b-a\right)\left(\frac{1}{\sqrt{b-4}+\sqrt{a-4}}+\frac{1}{\sqrt{b+1}+\sqrt{a+1}}\right)\)

\(\left\{{}\begin{matrix}b>a\Rightarrow b-a>0\\\frac{1}{\sqrt{b-4}+\sqrt{a-4}}+\frac{1}{\sqrt{b+1}+\sqrt{a+1}}>0\end{matrix}\right.\)

\(\Rightarrow y\left(b\right)-y\left(a\right)>0\) \(\forall b>a\)

\(\Rightarrow y\) đồng biến trên miền đã cho

Đáp án là D.

          • Sai ở bước III (bảng biến thiên)

a: ĐKXĐ: x∈R

Lấy x1,x2 sao cho x1<x2

=>x1-x2<0

\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2+2x_1+3}-\sqrt{x_2^2+2x_2+3}}{x_1-x_2}\)

\(=\frac{x_1^2+2x_1+3-x_2^2-2x_2-3}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+2x_1+3}+\sqrt{x_2^2+2x_2+3}\right)}\)

\(=\frac{x_1^2-x_2^2+2x_1-2x_2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+2x_1+3}+\sqrt{x_2^2+2x_2+3}\right)}\)

\(=\frac{\left(x_1-x_2)\left(x_1+x_2\right)+2\left(x_1-2x_2\right)\right.}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+2x_1+3}+\sqrt{x_2^2+2x_2+3}\right)}=\frac{x_1+x_2+2}{\sqrt{x_1^2+2x_1+3}+x_2^2+2x_2+3}\) >0

=>Hàm số đồng biến trên R

b: ĐKXĐ: \(x^2-3x+2\ge0\)

=>(x-1)(x-2)>=0

=>x>=2 hoặc x<=1

Lấy x1,x2 thuộc [2;+∞) sao cho 2<x1<x2

=>x1-2>0; x2-2>0

=>x1+x2-4>0

\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2-3x_1+2}-\sqrt{x_2^2-3x_2+2}}{x_1-x_2}\)

\(=\frac{x_1^2-3x_1+2-x_2^2+3x_2-2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{x_1^2-x_2^2-3x_1+3x_2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{\left(x_1^2-x_2^2\right)-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{\left(x_1-x_2\right)\left(x_1+x_2\right)_{}-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{\left(x_1+x_2\right)_{}-3}{\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}>0\)

=>Hàm số đồng biến trên khoảng [2;+∞)

Lấy x1,x2 thuộc (-∞;1] sao cho x2<x1<1

=>x1-1<0; x2-1<0

=>x1+x2-2<0

=>x1+x2-3<-1<0

\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2-3x_1+2}-\sqrt{x_2^2-3x_2+2}}{x_1-x_2}\)

\(=\frac{x_1^2-3x_1+2-x_2^2+3x_2-2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{x_1^2-x_2^2-3x_1+3x_2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{\left(x_1^2-x_2^2\right)-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{\left(x_1-x_2\right)\left(x_1+x_2\right)_{}-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)

\(=\frac{\left(x_1+x_2\right)_{}-3}{\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}>0\)

=>Hàm số nghịch biến trên khoảng (-∞;1]

+TXĐ: X\(\in\)R

+y'=\(3x^2-6x\Rightarrow y'=0\Leftrightarrow\int_{x=2;y=0}^{x=0;y=4}\)

+y''=6(x-1)=> y' = 0 khi x = 1;y=2

+

2.  y' = 3x² - 6x + m <0 khi x thuộc ( -1; 3)  => m/3 =-3 =>  m =-9

TXĐ: D=[0;+\(\infty\))

Hàm số này luôn đồng biến với mọi x thuộc D