Trên câu hỏi bn có ghi là tìm m để pt có gt max đâu mà mk biết ...

\(\Delta'=\left(m+1\right)^2-\left(5m+1\right)=m^2-3m\ge0\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=5m+1\end{matrix}\right.\)

a.

\(S=\left(x_1+x_2\right)^2-3x_1x_2=4\left(m+1\right)^2-3\left(5m+1\right)\)

\(=4m^2-7m+1=\dfrac{7}{3}\left(m^2-3m\right)+\dfrac{5}{3}m^2+1\ge1\)

\(S_{min}=1\) khi \(\dfrac{7}{3}\left(m^2-3m\right)+\dfrac{5}{3}m^2=0\Rightarrow m=0\)

b.

\(1< x_1< x_2\Rightarrow\left\{{}\begin{matrix}\left(x_1-1\right)\left(x_2-1\right)>0\\\dfrac{x_1+x_2}{2}>1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1x_2-\left(x_1+x_2\right)+1>0\\x_1+x_2>2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5m+1-2\left(m+1\right)+1>0\\2\left(m+1\right)>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m>0\\m>-1\end{matrix}\right.\) \(\Rightarrow m>0\)

Kết hợp điều kiện delta \(\Rightarrow m\ge3\)

\(a,\Leftrightarrow\Delta\ge0\Leftrightarrow\left(2m+2\right)^2-4\left(5m+1\right)\ge0\Leftrightarrow4m^2-12m\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}m\le0\\m\ge3\end{matrix}\right.\)

\(vi-ét\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1x2=5m+1\end{matrix}\right.\)

\(\Rightarrow S=x1^2+x2^2-x1x2=\left(x1+x2\right)^2-3x1x2\)

\(=\left(2m+2\right)^2-3\left(5m+1\right)=4m^2-7m+1\)

\(=\left(2m\right)^2-2.2.\dfrac{7}{4}.m+\left(\dfrac{7}{4}\right)^2-\dfrac{33}{16}=\left(2m-\dfrac{7}{4}\right)^2-\dfrac{33}{16}\left(1\right)\)

\(TH1:m\ge3\Rightarrow\left(1\right)\ge\left(2.3-\dfrac{7}{4}\right)^2-\dfrac{33}{16}=16\)

\(TH2:m\le0\Rightarrow\left(1\right)\ge\left(0-\dfrac{7}{4}\right)^2-\dfrac{33}{16}=1\)

\(\Rightarrow MinS=1\Leftrightarrow m=0\left(tm\right)\)

\(b,1< x1< x2\Leftrightarrow0< x1-1< x2-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left(x1-1\right)\left(x2-1\right)>0\\x1+x2-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x1>1\\x2>1\end{matrix}\right.\\\left\{{}\begin{matrix}x1 < 1\\x2< 1\end{matrix}\right.\end{matrix}\right.\\2m+2-2>0\\\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}x1x2>1\\x1x2< 1\end{matrix}\right.\\m>0\\\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< 0\end{matrix}\right.\\m>0\\\end{matrix}\right.\Rightarrow m>3\)

\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)

\(P=x_1x_2-\left(x_1^2+x_2^2\right)=3x_1x_2-\left(x_1+x_2\right)^2\)

\(P=3\left(m-2\right)-m^2=-m^2+3m-6=-\left(m-\dfrac{3}{2}\right)^2-\dfrac{15}{4}\le-\dfrac{15}{4}\)

\(P_{max}=-\dfrac{15}{4}\) khi \(m=\dfrac{3}{2}\)

\(P_{min}\) ko tồn tại

Bạn ghi sai đề?

\(Δ=(-m)^2-4.1.(m-2)\\=m^2-4m+8\\=m^2-4m+4+4\\=(m-2)^2+4\)

\(\to\) Pt luôn có 2 nghiệm phân biệt

Theo Viét

\(\begin{cases}x_1+x_2=m\\x_1x_2=m-2\end{cases}\)

\(x_1x_2-x_1^2-x_2^2\\=3x_1x_2-(x_1^2+2x_1x_2+x_2^2)\\=3x_1x_2-(x_1+x_2)^2\\=3(m-2)-m^2\\=-m^2+3m-6\\=-\bigg(m^2-2.\dfrac{3}{2}.m+\dfrac{9}{4}+\dfrac{15}{4}\bigg)\\=-\bigg(m-\dfrac{3}{2}\bigg)^2-\dfrac{15}{4}\le -\dfrac{15}{4}\\\to \max P=-\dfrac{15}{4}\leftrightarrow m-\dfrac{3}{2}=0\\\leftrightarrow m=\dfrac{3}{2}\)

Vậy \(\max P=-\dfrac{15}{4}\)

Ta có:

\(\Delta'=b'^2-ac=m^2-\left(2m-1\right)=m^2-2m+1=\left(m-1\right)^2\ge0\forall m\)

Vậy phương trình trên luôn có 2 nghiệm x₁; x₂ với mọi giá trị của m

Áp dụng Viet, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=-2m\\x_1\cdot x_2=\frac{c}{a}=2m-1\end{matrix}\right.\)

Ta có:

\(A=x_1^2\cdot x_2+x_1\cdot x_2^2\\ =x_1x_2\left(x_1+x_2\right)\\ =\left(2m-1\right)\cdot\left(-2m\right)\\ =-4m^2+2m\\ =-\left[\left(2m\right)^2-2\cdot2m\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+\frac{1}{4}\\ =-\left(2m-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall m\)

Vậy Max A = \(\frac{1}{4}\Leftrightarrow2m-\frac{1}{2}=0\Leftrightarrow m=\frac{1}{4}\left(tm\right)\)

Bài 2:

a: \(a=1;b=-2\left(m-2\right);c=-8\)

Vì ac<0 nên phương trình luôn có hai nghiệm trái dấu với mọi m

b: Theo Vi-et, ta được: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)=2m-4\\x_1x_2=-8\end{matrix}\right.\)

Ta có: \(x_1^3+x_2^3-4x_1-4x_2=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-4\left(x_1+x_2\right)=0\)

\(\Leftrightarrow\left(2m-4\right)^3-3\cdot\left(2m-4\right)\cdot\left(-8\right)-4\cdot\left(2m-4\right)=0\)

\(\Leftrightarrow\left(2m-4\right)\left[4m^2-16m+16+24-4\right]=0\)

\(\Leftrightarrow\left(2m-4\right)\left(4m^2-16m+36\right)=0\)

\(\Leftrightarrow2m-4=0\)

hay m=2

\(\Delta'=m^2-\left(2m^2-4m+3\right)=-m^2+4m-3\)

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

Để pt trên có 2 nghiệm x1 ; x2 khi \(0\le-\left(m-2\right)^2+1\le1\)

Theo Vi et : \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m^2-4m+3\end{matrix}\right.\)

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

\(=4m^2+2m^2-4m+3=6m^2-4m+4\)

bạn kiểm tra lại đề xem có vấn đề gì ko ? 

\(\Delta'=m^2-\left(2m^2-4m+3\right)=-m^2+4m-3\ge0\Rightarrow1\le m\le3\)

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m^2-4m+3\end{matrix}\right.\)

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

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

\(=6m^2-4m+3\)

Xét hàm \(f\left(m\right)=6m^2-4m+3\) trên \(\left[1;3\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{3}< 1;a=6>0\Rightarrow f\left(m\right)\) đồng biến trên \(\left[1;3\right]\)

\(\Rightarrow f\left(m\right)_{max}=f\left(3\right)=45\) khi \(m=3\)

1) Thay m=1 vào phương trình, ta được:

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

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

\(\Leftrightarrow x-1=0\)

hay x=1

Vậy: Khi m=1 thì phương trình có nghiệm duy nhất là x=1

1) Bạn tự làm

2) Ta có: \(\Delta'=\left(m-1\right)^2\ge0\)

\(\Rightarrow\) Phương trình luôn có 2 nghiệm

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m-1\end{matrix}\right.\) 

a) Ta có: \(x_1+x_2=-1\) \(\Rightarrow2m=-1\) \(\Leftrightarrow m=-\dfrac{1}{2}\)

   Vậy ...

b) Ta có: \(x_1^2+x_2^2=13\) \(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=13\)

            \(\Rightarrow4m^2-4m-11=0\) \(\Leftrightarrow m=\dfrac{1\pm\sqrt{13}}{2}\)

  Vậy ... 

a) ta có :

\(\Delta'\) = (-m)² +1 = m² + 1 \(\ge\) 1

=> pt luôn có 2 nghiệm phân biệt x₁ , x₂

b) theo hệ thức vi - ét thì : \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1.x_2=-1\end{matrix}\right.\)

ta có : x₁² + x₂² - x₁x₂ = 7

<=> x₁² + 2x₁x₂ + x₂² - 2x₁x₂ - x₁x₂ = 7

<=> ( x₁ + x₂ )² - 3x₁x₂ = 7

<=> (2m)² - 3.(-1) = 7

<=> 4m² +3 = 7

<=> 4m² = 4

<=> m² = 1

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