2: Tọa độ A là:

\(\left\{{}\begin{matrix}x=0\\y=\left(m-2\right)x+m-1=0\left(m-2\right)+m-1=m-1\end{matrix}\right.\)

=>A(0;m-1)

Tọa độ B là:

\(\left\{{}\begin{matrix}y=0\\\left(m-2\right)x+m-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x\left(m-2\right)=-\left(m-1\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{-\left(m-1\right)}{m-2}\\y=0\end{matrix}\right.\)

Vậy: \(B\left(\dfrac{-m+1}{m-2};0\right)\)

\(OA=\sqrt{\left(0-0\right)^2+\left(m-1-0\right)^2}=\sqrt{0+\left(m-1\right)^2}=\sqrt{\left(m-1\right)^2}=\left|m-1\right|\)

\(OB=\sqrt{\left(\dfrac{-m+1}{m-2}-0\right)^2+\left(0-0\right)^2}\)

\(=\sqrt{\left(-\dfrac{m-1}{m-2}\right)^2+0}=\left|\dfrac{m-1}{m-2}\right|\)

Vì Ox\(\perp\)Oy

nên OA\(\perp\)OB

=>\(S_{OAB}=\dfrac{1}{2}\cdot\left|m-1\right|\cdot\dfrac{\left|m-1\right|}{\left|m-2\right|}=\dfrac{1}{2}\cdot\dfrac{\left(m-1\right)^2}{\left|m-2\right|}\)

Để \(S_{OAB}=1\) thì \(\dfrac{1}{2}\cdot\dfrac{\left(m-1\right)^2}{\left|m-2\right|}=1\)

=>\(\left(m-1\right)^2=2\left|m-2\right|\)(1)

TH1: m>=2

Phương trình (1) sẽ trở thành: \(\left(m-1\right)^2=2\left(m-2\right)\)

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

=>\(m^2-4m+5=0\)

=>\(\left(m-2\right)^2+1=0\)(vô lý)

TH2: m<2

Phương trình (1) sẽ trở thành:

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

=>\(m^2-2m+1=-2m+4\)

=>m²=3

=>\(\left[{}\begin{matrix}m=\sqrt{3}\left(nhận\right)\\m=-\sqrt{3}\left(nhận\right)\end{matrix}\right.\)

a. \(f\left(x\right)_{max}=f\left(-2\right)=111\) ; \(f\left(x\right)_{min}=f\left(1\right)=-6\)

b. \(f\left(x\right)_{max}=f\left(-3\right)=7\) ; \(f\left(x\right)_{min}=f\left(0\right)=1\)

c. \(f\left(x\right)_{max}=f\left(4\right)=\dfrac{2}{3}\) ; \(f\left(x\right)_{min}\) ko tồn tại

d. 

Miền xác định: \(D=\left[-2\sqrt{2};2\sqrt{2}\right]\)

\(y'=\dfrac{2\left(4-x^2\right)}{\sqrt{8-x^2}}=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

\(f\left(-2\sqrt{2}\right)=f\left(2\sqrt{2}\right)=0\)

\(f\left(-2\right)=-4\) ; \(f\left(2\right)=4\)

\(f\left(x\right)_{max}=f\left(2\right)=4\) ; \(f\left(x\right)_{min}=f\left(-2\right)=-4\)

11 c)

\(a^2+2\ge2\sqrt{a^2+1}\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\Leftrightarrow\left(\sqrt{a^2+1}-1\right)^2\ge0\) (luôn đúng)

12 a)  Có a+b+c=1\(\Rightarrow\) (1-a)(1-b)(1-c)= (b+c)(a+c)(a+b) (*)

áp dụng BĐT cô-si: \(\left(b+c\right)\left(a+c\right)\left(a+b\right)\ge2\sqrt{bc}2\sqrt{ac}2\sqrt{ab}=8\sqrt{\left(abc\right)2}=8abc\) ( luôn đúng với mọi a,b,c ko âm ) 

b)  áp dụng BĐT cô-si: \(c\left(a+b\right)\le\dfrac{\left(a+b+c\right)^2}{4}=\dfrac{1}{4}\)

Tương tự: \(a\left(b+c\right)\le\dfrac{1}{4};b\left(c+a\right)\le\dfrac{1}{4}\)

\(\Rightarrow abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{4}\dfrac{1}{4}\dfrac{1}{4}=\dfrac{1}{64}\)

a: \(\Delta=\left(-m\right)^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>\(\left(m-2\right)^2>0\)

=>m-2<>0

=>m<>2

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=m;x_1x_2=\frac{c}{a}=m-1\)

\(x_1-x_2=5\)

=>\(\left(x_1-x_2\right)^2=5^2=25\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2=25\)

=>\(m^2-4\left(m-1\right)=25\)

=>\(m^2-4m+4=25\)

=>\(m^2-4m-21=0\)

=>(m-7)(m+3)=0

=>m=7(nhận) hoặc m=-3(nhận)

b: \(\frac{1}{x_1-2}+\frac{1}{x_2-2}=\frac12\)

=>\(\frac{x_2-2+x_1-2}{\left(x_1-2\right)\left(x_2-2\right)}=\frac12\)

=>\(\frac{x_1+x_2-4}{x_1x_2-2\left(x_1+x_2\right)+4}=\frac12\)

=>\(\frac{m-4}{m-1-2m+4}=\frac12\)

=>\(\frac{m-4}{-m+3}=\frac12\)

=>2(m-4)=-m+3

=>2m-8=-m+3

=>3m=11

=>\(m=\frac{11}{3}\) (nhận)

c: \(\left|x_1\right|=2\left|x_2\right|\)

=>\(\left[\begin{array}{l}x_1=2x_2\\ x_1=-2x_2\end{array}\right.\)

TH1: \(x_1=2x_2\)

mà \(x_1+x_2=m\)

nên \(x_1=\frac{2m}{3};x_2=\frac{m}{3}\)

\(x_1\cdot x_2=m-1\)

=>\(\frac{2m}{3}\cdot\frac{m}{3}=m-1\)

=>\(2m^2=9\left(m-1\right)=9m-9\)

=>\(2m^2-9m+9=0\)

=>\(2m^2-3m-6m+9=0\)

=>m(2m-3)-3(2m-3)=0

=>(2m-3)(m-3)=0

=>\(\left[\begin{array}{l}m=\frac32\left(nhận\right)\\ m=3\left(nhận\right)\end{array}\right.\)

TH2: \(x_1=-2x_2\)

\(x_1+x_2=m\)

=>\(-2x_2+x_2=m\)

=>\(-x_2=m\)

=>\(x_2=-m\)

=>\(x_1=-2\cdot\left(-m\right)=2m\)

\(x_1x_2=m-1\)

=>\(-2m^2=m-1\)

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

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

=>(m+1)(2m-1)=0

=>m=-1(nhận) hoặc m=1/2(nhận)

d: \(P=x_1^2+x_2^2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=m^2-2\left(m-1\right)=m^2-2m+2=m^2-2m+1+1=\left(m-1\right)^2+1\ge1\forall m\)

Dấu '=' xảy ra khi m-1=0

=>m=1

là vui đố ngày xuân ạk

enough strong => strong enough nhé bạn, adj + enough / enough + noun