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

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

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

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

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

Xét phương trình \(x^2-mx+1005m=0\) có \(\Delta=m^2-4.1005m=m^2-4020m\)

Do pt có hai nghiệm nên \(\Delta\ge0\Leftrightarrow\left[{}\begin{matrix}m\le0\\m\ge4020\end{matrix}\right.\)

Theo hệ thức Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=1005m\end{matrix}\right.\)

\(\Rightarrow M=\dfrac{2.1005m+2680}{m^2+1}=\dfrac{2010m+2680}{m^2+1}\)

\(=335\left(\dfrac{\left(m+3\right)^2}{m^2+1}-1\right)\ge-335\)

Vậy minM = -335, khi m = -3.

\(\ast\Delta>0\Leftrightarrow m^2-4.1005m>0\Leftrightarrow m<0\text{ hoặc }m>4020\)

\(\ast x_1.x_2=1005m;\text{ }x_1+x_2=m\)

\(P=\frac{2x_1.x_2+2680}{\left(x_1+x_2\right)^2+1}=\frac{2010m+2680}{m^2+1}=670.\frac{3m+4}{m^2+1}\)

\(=670.\left(\frac{3m+4}{m^2+1}+\frac{1}{2}\right)-\frac{670}{2}=670.\frac{m^2+1+2\left(3m+4\right)}{2\left(m^2+1\right)}-335\)

\(=335.\frac{\left(m+3\right)^2}{m^2+1}-335\ge-335\)

Dấu bằng xảy ra khi \(m+3=0\Leftrightarrow m=-3\text{ }\left(\text{thỏa}\right)\)

Vậy \(m=-3\)

xét pt \(x^2-mx+m-1=0\)  \(\left(1\right)\)

xó \(\Delta=\left(-m\right)^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2>0\forall m\ne2\)

\(\Rightarrow pt\)  (1) có 2 nghiệm phân biệt \(x_1,x_2\forall m\ne2\)

ta có vi -ét \(\hept{\begin{cases}x_1+x_2=m\\x_1.x_2=m-1\end{cases}}\)

theo bài ra \(\left|x_1\right|+\left|x_2\right|=6\)

\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=36\)

\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1.x_2\right|=36\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=36\)

\(\Leftrightarrow m^2-2\left(m-1\right)+2\left|m-1\right|=36\)

nếu \(m-1< 0\Rightarrow m^2-4m-32=0\)  ta tìm được \(m=8\left(loai\right)\); \(m=-4\left(TM\right)\)

nếu \(m-1\ge0\Rightarrow m^2=36\Rightarrow m=6\left(TM\right);m=-6\left(loai\right)\)

vậy \(m=-4;m=6\)  là các giá trị cần tìm 

b) \(P=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2-2x_1x_2+2x_1x_2+2}\)

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

\(P=\frac{2m-2+3}{m^2+2}=\frac{2m+1}{m^2+2}\)

vậy \(P=\frac{2m+1}{m^2+2}\)

c) \(P=\frac{2m+1}{m^2+2}=\frac{m^2+2-m^2+2m-1}{m^2+2}=1-\frac{m^2-2m+1}{m^2+2}\)

\(P=1-\frac{\left(m-1\right)^2}{m^2+2}\le1\)

dấu \("="\)  xảy ra \(\Leftrightarrow m-1=0\Leftrightarrow m=1\)

vậy \(MAX\)   \(P=1\Leftrightarrow m=1\)

\(P=\frac{2m+1}{m^2+2}=\frac{4m+2}{2\left(m^2+2\right)}=\frac{m^2+4m+4-m^2-2}{2\left(m^2+2\right)}\)

\(P=\frac{\left(m+2\right)^2-m^2-2}{2\left(m^2+2\right)}=\frac{\left(m+2\right)^2}{2\left(m^2+2\right)}-\frac{m^2+2}{2\left(m^2+2\right)}\)

\(P=\frac{\left(m+2\right)^2}{2\left(m^2+2\right)}-\frac{1}{2}\ge\frac{1}{2}\)

dấu \("="\)  xảy ra \(\Leftrightarrow m+2=0\Leftrightarrow m=-2\)

vậy \(MIN\)   \(P=\frac{1}{2}\Leftrightarrow m=-2\)

Bạn xem lại đề bài

\(2m^2-2mx....\) có gì đó sai sai

b/ Theo vi - et thì:

\(\hept{\begin{cases}x_1+x_2=m\\x_1x_2=m-1\end{cases}}\)

Ta có:

\(A=\frac{1}{x^2_1x_2+\left(m-1\right)x_2+1}-\frac{4}{x_1x^2_2+\left(m-1\right)x_1+1}\)

\(=\frac{1}{\left(m-1\right)x_1+\left(m-1\right)x_2+1}-\frac{4}{\left(m-1\right)x_2+\left(m-1\right)x_1+1}\)

\(=\frac{1}{m\left(m-1\right)+1}-\frac{4}{m\left(m-1\right)+1}\)

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

\(\ge-\frac{3}{\frac{3}{4}}=-4\)

Vậy GTNN là A = - 4 đạt được khi \(m=\frac{1}{2}\) 

Em không hiểu dòng 2 của biểu thức ý..

Dòng 2 a thay x1x2 = m - 1 đó e. Thay vô là thấy liền ah