\(\Delta'=m^2-\left(m^2+2m-6\right)=-2m+6\)

a.

Pt có nghiệm khi \(-2m+6\ge0\Rightarrow m\le3\)

b.

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2+2m-6\end{matrix}\right.\)

c.

\(x_1x_2=3x_1+3x_2-1\)

\(\Leftrightarrow x_1x_2=3\left(x_1+x_2\right)-1\)

\(\Leftrightarrow m^2+2m-6=3.2m-1\)

\(\Leftrightarrow m^2-4m-5=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=5>3\left(loại\right)\end{matrix}\right.\)

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

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

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

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

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

a) Ta có :  \(\Delta"=\left(-m\right)^2-\left(m-2\right)=m^2-m+2=\left(m-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall m\)

=> Phương trình luôn có 2 nghiệm phân biệt

b) Hệ thức Viete : 

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

Khi đó \(M=\dfrac{-24}{x_1^2+x_2^2-6x_1x_2}=\dfrac{-24}{\left(x_1+x_2\right)^2-8x_1x_2}\)

\(=\dfrac{-24}{\left(2m\right)^2-8.\left(m-2\right)}=\dfrac{-6}{m^2-2m+4+=}=\dfrac{-6}{\left(m-1\right)^2+3}\)

Do (m - 1)² + 3 \(\ge3\forall m\)

nên \(\dfrac{6}{\left(m-1\right)^2+3}\le2\Leftrightarrow M=\dfrac{-6}{\left(m-1\right)^2+3}\ge-2\)

Vậy M_min = -2 <=> m = 1

a, b bạn tự giải

c. \(\Delta=m^2+4>0;\forall m\Rightarrow\) pt luôn có nghiệm

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

Ồ, đề câu d bạn ghi sai, 2 mẫu số phải có 1 cái là \(x_1\)

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

      \(=16-4+8m=12+8m\)

Để pt có 2 nghiệm thì \(12+8m>0\)

                                       \(\Leftrightarrow m>-\dfrac{12}{8}\)

b. Theo hệ thức vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1.x_2=1-2m\end{matrix}\right.\)

\(x_1^2+x^2_2=6\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=6\)

\(\Leftrightarrow4^2-2\left(1-2m\right)=6\)

\(\Leftrightarrow16-2+4m-6=0\)

\(\Leftrightarrow4m=-8\)

\(\Leftrightarrow m=-2\)

a, \(\Delta'=\left(-2\right)^2-\left(1-2m\right)=4-1+2m=2m-3\)

Để pt có nghiệm thì \(\Delta'\ge0\Leftrightarrow2m-3\ge0\Leftrightarrow m\ge\dfrac{3}{2}\)

b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=1-2m\end{matrix}\right.\)

\(x_1^2+x_2^2=6\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=6\\ \Leftrightarrow4^2-2\left(1-2m\right)=6\\ \Leftrightarrow16-2+4m-6=0\\ \Leftrightarrow4m-8=0\\ \Leftrightarrow m=2\left(tm\right)\)

\(a,m=1\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

\(b,\) PT có 2 nghiệm pb \(\Leftrightarrow\Delta=4\left(m+1\right)^2-4\left(m^2+2\right)>0\\ \Leftrightarrow4m^2+8m+4-4m^2-8>0\\ \Leftrightarrow8m-4>0\Leftrightarrow m>\dfrac{1}{2}\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{matrix}\right.\)

Ta có \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=10\)

\(\Leftrightarrow4\left(m+1\right)^2-2\left(m^2+2\right)=10\\ \Leftrightarrow4m^2+8m+4-2m^2-4=10\\ \Leftrightarrow2m^2+8m-10=0\\ \Leftrightarrow m^2+4m-5=0\\ \Leftrightarrow\left(m+5\right)\left(m-1\right)=0\Leftrightarrow m=1\left(m>\dfrac{1}{2}\right)\)

Vậy m=1 thỏa mãn đề bài

\(\Delta=1-4m>0\Rightarrow m< \dfrac{1}{4}\)

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

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

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

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

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

\(\Leftrightarrow m^2-m-2=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=2>\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)