a, Phương trình có hai nghiệm khi 

\(\Delta'=m^2-2\left(m^2-2\right)=-m^2+4\ge0\Leftrightarrow-2\le m\le2\)

b, Theo định lí Viet \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=\dfrac{m^2-2}{2}\end{matrix}\right.\)

\(A=\left|2x_1x_2+x_1+x_2-4\right|\)

\(=\left|m^2-2-m-4\right|\)

\(=\left|\left(m-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\right|\)

\(=\left|-\left(m-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\right|\le\dfrac{25}{4}\)

\(maxA=\dfrac{25}{4}\Leftrightarrow m=\dfrac{1}{2}\)

Bạn xem lại đề bài

\(2m^2-2mx....\) có gì đó sai sai

\(\Delta'=\left(m-1\right)^2-\left(-2m+5\right)=m^2-4\ge0\Rightarrow\left[{}\begin{matrix}m\ge2\\m\le-2\end{matrix}\right.\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-1\right)\\x_1x_2=-2m+5\end{matrix}\right.\)

a/ Kết hợp Viet và đề bài ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-1\right)\\2x_1+3x_2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x_1+3x_2=-6\left(m-1\right)\\2x_1+3x_2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-6m+5\\x_2=4m-3\end{matrix}\right.\)

\(\Rightarrow\left(-6m+5\right)\left(4m-3\right)=-2m+5\)

\(\Leftrightarrow-24m^2+38m-15=-2m+5\)

\(\Leftrightarrow24m^2-40m+20=0\)

Phương trình vô nghiệm \(\Rightarrow\) ko tồn tại m thỏa mãn yêu cầu

b/

Ta có:

\(A=12-10x_1x_2-\left(x_1^2+x_2^2\right)\)

\(A=12-10x_1x_2-\left(x_1+x_2\right)^2+2x_1x_2\)

\(A=12-8x_1x_2-\left(x_1+x_2\right)^2\)

\(A=12-8\left(-2m+5\right)-4\left(m-1\right)^2\)

\(A=12+16m-40-4m^2+8m-4\)

\(A=-4m^2+24m-32=0\)

\(A=-4\left(m-3\right)^2+4\le4\)

\(\Rightarrow A_{max}=4\) khi \(m=3\)

c/

Ta có:

\(x_1+x_2+2x_1x_2\le6\)

\(\Leftrightarrow-2\left(m-1\right)+2\left(-2m+5\right)\le6\)

\(\Leftrightarrow-2m+2-4m+10\le6\)

\(\Leftrightarrow-6m\le-6\)

\(\Rightarrow m\ge1\)

Kết hợp với điều kiện \(\Delta\) ta có: \(m\ge2\)

b) phương trình có 2 nghiệm  \(\Leftrightarrow\Delta'\ge0\)

\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)

\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)

\(\Leftrightarrow-4m+4\ge0\)

\(\Leftrightarrow m\le1\)

Ta có: \(x_1^2+x_1x_2+x_2^2=1\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)

\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)

\(\Leftrightarrow4m^2-10m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)

Câu c:

\(\Delta^`\ge0\)

\(\Leftrightarrow m^2-\left(m^2-2\right).2\ge0\)

\(\Leftrightarrow4-m^2\ge0\)

\(\Leftrightarrow4\ge m^2\)

\(\Leftrightarrow4\ge m^2\)

\(\Leftrightarrow-2\le m\le2\)

Theo hệ thức Viet có:

\(\hept{\begin{cases}x_1+x_2=m\\x_1.x_2=\frac{m^2-2}{2}\end{cases}}\)

\(\Rightarrow A=\left|2x_1.x_2-x_1-x_2-4\right|=\left|m^2-m-6\right|=\left|\left(m-\frac{1}{2}\right)^2-6,25\right|\)

Có:

\(\left(m-\frac{1}{2}\right)^2\le\left(-2-\frac{1}{2}\right)^2=6,25\)

\(\Rightarrow A=\left|\left(m-\frac{1}{2}\right)^2-6,25\right|=6,25-\left(m-\frac{1}{2}\right)^2\le6,25\)

\(A=6,25\Leftrightarrow m=\frac{1}{2}\left(tm\right)\)

KL:..............................................

1.

\(a+b+c=0\) nên pt luôn có 2 nghiệm

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

\(A=\dfrac{2x_1x_2+3}{x_1^2+x_2^2+2x_1x_2+2}=\dfrac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\dfrac{2\left(m-1\right)+3}{m^2+2}=\dfrac{2m+1}{m^2+2}\)

\(A=\dfrac{m^2+2-\left(m^2-2m+1\right)}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)

Dấu "=" xảy ra khi \(m=1\)

2.

\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\) nên pt luôn có 2 nghiệm pb

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

\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)

\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)

\(m>1\Rightarrow ac=-m-3< 0\Rightarrow\) pt luôn có 2 nghiệm trái dấu

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m-3\end{matrix}\right.\)

\(A=\dfrac{2\left(x_1+x_2\right)^2-6x_1x_2}{x_1+x_2}=\dfrac{2.4\left(m-1\right)^2+6\left(m+3\right)}{2\left(m-1\right)}\)

\(=\dfrac{4\left(m-1\right)^2+3\left(m-1\right)+12}{m-1}=4\left(m-1\right)+\dfrac{12}{m-1}+3\)

\(A\ge2\sqrt{4\left(m-1\right).\dfrac{12}{m-1}}+3=3+8\sqrt{3}\)

Dấu "=" xảy ra khi \(4\left(m-1\right)=\dfrac{12}{m-1}\Rightarrow m=1+\sqrt{3}\)