b: \(PT\Leftrightarrow x^2+\left(m-3\right)x-m=0\)

\(\text{Δ}=\left(m-3\right)^2+4m\)

\(=m^2-6m+9+4m\)

\(=m^2-2m+1+8=\left(m-1\right)^2+8>0\)

Do đó: PT luon có hai nghiệm phân biệt

\(\dfrac{2}{x_1}+\dfrac{2}{x_2}=\dfrac{2x_1+2x_2}{x_1x_2}=\dfrac{2\cdot\left(-m+3\right)}{-m}=\dfrac{-2m+6}{-m}\)

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

\(=\dfrac{4\left(x_1+x_2\right)^2-8x_1x_2}{x_1x_2}=\dfrac{4\left(-m+3\right)^2-8\cdot\left(-m\right)}{-m}\)

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

\(=\dfrac{4m^2-24m+36+8m}{-m}=\dfrac{4m^2-16m+36}{-m}\)

c: \(A=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}+1\)

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

\(=\sqrt{m^2-6m+9+4m}+1\)

\(=\sqrt{m^2-2m+1+8}+1\)

\(=\sqrt{\left(m-1\right)^2+8}+1\ge2\sqrt{2}+1\)

Dấu '=' xảy ra khi m=1

Theo vi-et thì ta có:

\(\hept{\begin{cases}x_1+x_2=\frac{3a-1}{2}\\x_1x_2=-1\end{cases}}\)

Từ đây ta có: 

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=\left(\frac{3a-1}{2}\right)^2-4.1=\left(\frac{3a-1}{2}\right)^2-4\)

Theo đề bài thì 

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

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

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

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

\(=6\left(\left(\frac{3a-1}{2}\right)^2-4\right)\ge6.4=24\)

Dấu = xảy ra khi \(a=\frac{1}{3}\)

\(x^2-2\left(m+1\right)x+3\left(m+1\right)-3=0\)

\(x^2-2nx+3n+3=\left(x-n\right)^2-\left(n^2-3n+3\right)=0\)\(\left(x-n\right)^2=\left(n-\frac{3}{2}\right)^2+\frac{3}{4}=\frac{\left(2n-3\right)^2+3}{4}>0\forall n\) vậy luôn tồn tại hai nghiệm

\(\orbr{\begin{cases}x_1=\frac{n-\sqrt{\left(2n-3\right)^2+3}}{2}\\x_2=\frac{n+\sqrt{\left(2n-3\right)^2+3}}{2}\end{cases}}\)

a) \(\frac{x_1}{x_2}=\frac{4x_1-x_2}{x_1}\Leftrightarrow\frac{x_1^2-4x_1x_2+x_2^2}{x_1x_2}=0\)

\(x_1x_2=n^2-\frac{\left(2n-3\right)^2+3}{4}=\frac{4n^2-4n^2+12n-9-3}{4}=3n-3\)

với n=1 hay m=0 : Biểu thức cần C/m không tồn tại => xem lại đề

\(x^2-x\left(m+2\right)-m-3=0\)

Ta có \(\Delta=\left(m+2\right)^2-4\cdot\left(-m-3\right)\ge0\)

\(\Leftrightarrow m^2+8m+16\ge0\)

\(\Leftrightarrow\left(m+4\right)^2\ge0\) ( luôn đúng )

Theo định lý Vi-ét ta có:

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

Khi đó \(A=-x_1^2x_2-x_1x_2^2=-x_1x_2\left(x_1+x_2\right)=\left(m+2\right)\left(m+3\right)\)

\(A=m^2+5m+6\)

\(A=\left(m+\frac{5}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall m\)

Vậy \(A_{min}=\frac{-1}{4}\Leftrightarrow m=\frac{-5}{2}\)

Theo hệ thức Vi ét ,ta có:

\(\hept{\begin{cases}x_1+x_2=2m\\x_1\cdot x_2=2m-1\end{cases}}\)

\(2\left(x_1^2+x_2^2\right)-5x_1x_2=27\Leftrightarrow2\left(x_1^2+x_2^2+2x_1x_2\right)-9x_1x_2=27\)

\(2\left(x_1+x_2\right)^2-9x_1x_2=27\)

\(\Rightarrow2\left(2m\right)^2-9\left(2m-1\right)=27\\ \Leftrightarrow8m^2-18m+9=0\)

\(\Rightarrow\orbr{\begin{cases}m=\frac{3}{2}\\m=\frac{3}{4}\end{cases}}\)

\(\Delta'=\left(m+1\right)^2-\left(2m-3\right)=m^2+4>0,\forall m\inℝ\)

nên phương trình luôn có hai nghiệm phân biệt \(x_1+x_2\). 

Theo định lí Viete: 

\(\hept{\begin{cases}x_1+x_2=2m+2\\x_1x_2=2m-3\end{cases}}\)

\(P=\left|\frac{x_1+x_2}{x_1-x_2}\right|=\frac{\left|x_1+x_2\right|}{\left|x_1-x_2\right|}=\frac{\left|x_1+x_2\right|}{\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}}\)

\(=\frac{\left|2m+2\right|}{\sqrt{\left(2m+2\right)^2-4\left(2m-3\right)}}=\frac{\left|2m+2\right|}{\sqrt{4m^2+16}}=\frac{\left|m+1\right|}{\sqrt{m^2+4}}\ge0\)

Dấu \(=\)xảy ra khi \(m=-1\).