a: \(\text{Δ}=\left(2m+1\right)^2-4m\left(m+3\right)\)

\(=4m^2+4m+1-4m^2-12m\)

\(=-8m+1\)

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

\(\Leftrightarrow-8m+1>0\)

\(\Leftrightarrow-8m>-1\)

hay \(m< \dfrac{1}{8}\)

|x1|=3|x2|

=>|2m+2-x2|=|3x2|

=>4x2=2m+2 hoặc -2x2=2m+2

=>x2=1/2m+1/2 hoặc x2=-m-1

Th1: x2=1/2m+1/2

=>x1=2m+2-1/2m-1/2=3/2m+3/2

x1*x2=m^2+2m

=>1/2(m+1)*3/2(m+1)=m^2+2m

=>3/4m^2+3/2m+3/4-m^2-2m=0

=>m=1 hoặc m=-3

TH2: x2=-m-1 và x1=2m+2+m+1=3m+3

x1x2=m^2+2m

=>-3m^2-6m-3-m^2-2m=0

=>m=-1/2; m=-3/2

a. Phương trình có 2 nghiệm phân biệt khi:

\(\Delta=\left(2m-1\right)^2-4\left(m^2-1\right)=5-4m>0\)

\(\Rightarrow m< \dfrac{5}{4}\)

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

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

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=x_1-3x_2\)

\(\Leftrightarrow\left(2m-1\right)^2-4\left(m^2-1\right)=x_1-3x_2\)

\(\Leftrightarrow x_1-3x_2=5-4m\)

Kết hợp hệ thức Viet ta được: \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1-3x_2=5-4m\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2m-1\\4x_2=6m-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{m+1}{2}\\x_2=\dfrac{3m-3}{2}\end{matrix}\right.\)

Thế vào \(x_1x_2=m^2-1\)

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

\(\Leftrightarrow m^2-1=0\Rightarrow m=\pm1\) (thỏa mãn)

`1)`

$a\big)\Delta=7^2-5.4.1=29>0\to$ PT có 2 nghiệm pb

$b\big)$

Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{7}{5}\\x_1x_2=\dfrac{1}{5}\end{matrix}\right.\)

\(A=\left(x_1-\dfrac{7}{5}\right)x_1+\dfrac{1}{25x_2^2}+x_2^2\\ \Rightarrow A=\left(x_1-x_1-x_2\right)x_1+\left(\dfrac{1}{5}\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\\ \Rightarrow A=-x_1x_2+\left(x_1x_2\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\)

\(\Rightarrow A=-x_1x_2+x_1^2+x_2^2\\ \Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2\\ \Rightarrow A=\left(\dfrac{7}{5}\right)^2-3\cdot\dfrac{1}{5}=\dfrac{34}{25}\)

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

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

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

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

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

Đặt \(x^2=t\) \(\Rightarrow t^2+\left(1-m\right)t+2m-2=0\) (1)

Pt đã cho có 4 nghiệm pb \(\Leftrightarrow\) (1) có 2 nghiệm dương pb

\(\Rightarrow\left\{{}\begin{matrix}\Delta=\left(1-m\right)^2-8\left(m-1\right)>0\\t_1+t_2=m-1>0\\t_1t_2=2m-2>0\end{matrix}\right.\) \(\Rightarrow m>9\)

Khi đó, do vai trò của \(x_1;x_2;x_3;x_4\) như nhau, ko mất tính tổng quát, giả sử \(x_1=-\sqrt{t_1};x_2=\sqrt{t_1}\) ; \(x_3=-\sqrt{t_2};x_4=\sqrt{t_2}\)

\(\Rightarrow x_1x_2x_3x_4=t_1t_2\) ; \(x_1^2=x_2^2=t_1\) ; \(x_3^2=x_4^2=t_2\)

\(\Rightarrow\dfrac{x_1x_2x_3x_4}{2x_4^2}+\dfrac{x_1x_2x_3x_4}{2x_3^2}+\dfrac{x_1x_2x_3x_4}{2x_2^2}+\dfrac{x_1x_2x_3x_4}{2x_1^2}=2017\)

\(\Leftrightarrow\dfrac{t_1t_2}{2t_2}+\dfrac{t_1t_2}{2t_2}+\dfrac{t_1t_2}{2t_1}+\dfrac{t_1t_2}{2t_1}=2017\)

\(\Leftrightarrow t_1+t_2=2017\)

\(\Leftrightarrow m-1=2017\Rightarrow m=2018\)

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

\(\text{Δ}=\left(-2m+2\right)^2-4\cdot1\cdot\left(-2m\right)\)

\(=4m^2-8m+4+8m=4m^2+4>=4>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

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

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

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

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

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

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

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

Δ=(2m-2)^2-4(m-3)

=4m^2-8m+4-4m+12

=4m^2-12m+16

=4m^2-12m+9+7=(2m-3)^2+7>=7>0 với mọi m

=>Phương trình luôn có hai nghiệm phân biệt

\(\left(\dfrac{1}{x1}-\dfrac{1}{x2}\right)^2=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}-\dfrac{2}{x_1x_2}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{\left(\left(x_1+x_2\right)^2-2x_1x_2\right)}{\left(x_1\cdot x_2\right)^2}-\dfrac{2}{x_1\cdot x_2}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{\left(2m-2\right)^2-2\left(m-3\right)}{\left(-m+3\right)^2}-\dfrac{2}{-m+3}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{4m^2-8m+4-2m+6}{\left(m-3\right)^2}+\dfrac{2}{m-3}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{4m^2-10m+10+2m-6}{\left(m-3\right)^2}=\dfrac{\sqrt{11}}{2}\)

=>\(\sqrt{11}\left(m-3\right)^2=2\left(4m^2-8m+4\right)\)

=>\(\sqrt{11}\left(m-3\right)^2=2\left(2m-2\right)^2\)

=>\(\Leftrightarrow\left(\dfrac{m-3}{2m-2}\right)^2=\dfrac{2}{\sqrt{11}}\)

=>\(\left[{}\begin{matrix}\dfrac{m-3}{2m-2}=\sqrt{\dfrac{2}{\sqrt{11}}}\\\dfrac{m-3}{2m-2}=-\sqrt{\dfrac{2}{\sqrt{11}}}\end{matrix}\right.\)

mà m nguyên

nên \(m\in\varnothing\)