a) Có: `\Delta'=(m-2)^2-(m^2-4m)=m^2-4m+4-m^2+4m=4>0 forall m`

`=>` PT luôn có 2 nghiệm phân biệt với mọi `m`.

b) Viet: `x_1+x_2=-2m+4`

`x_1x_2=m^2-4m`

`3/(x_1) + x_2=3/(x_2)+x_1`

`<=> 3x_2+x_1x_2^2=3x_1+x_1^2 x_2`

`<=> 3(x_1-x_2)+x_1x_2(x_1-x_2)=0`

`<=>(x_1-x_2).(3+x_1x_2)=0`

`<=> \sqrt((x_1+x_2)^2-4x_1x_2) .(3+x_1x_2)=0`

`<=> \sqrt((-2m+4)^2-4(m^2-4m)) .(3+m^2-4m)=0`

`<=>  4.(3+m^2-4m)=0`

`<=> m^2-4m+3=0`

`<=>` \(\left[{}\begin{matrix}m=3\\m=1\end{matrix}\right.\)

Vậy `m \in {1;3}`.

\(ac=-3< 0\Rightarrow\) pt đã cho luôn có 2 nghiệm pb trái dấu với mọi m

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

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

\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}{9}=m-1\)

\(\Leftrightarrow8\left(m-1\right)^3+18\left(m-1\right)=9\left(m-1\right)\)

\(\Leftrightarrow\left(m-1\right)\left[8\left(m-1\right)^2+9\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\8\left(m-1\right)^2+9=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Delta=5^2-4\left(m-2\right)=25-4m+8=33-4m\)

Để pt có 2 nghiệm thì \(\Delta\ge0\Leftrightarrow33-4m\ge0\Leftrightarrow m\le\dfrac{33}{4}\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=m-2\end{matrix}\right.\)

\(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}=2\\ \Leftrightarrow\dfrac{x_2-1+x_1-1}{\left(x_1-1\right)\left(x_2-1\right)}=2\\ \Leftrightarrow x_1+x_2-2=2\left(x_1x_2-x_2-x_1+1\right)\\ \Leftrightarrow-5-2=2x_1x_2-2\left(x_1+x_2\right)+2\\ \Leftrightarrow2\left(m-2\right)-2.\left(-5\right)+2+7=0\\ \Leftrightarrow2m-4+10+2+7=0\\ \Leftrightarrow2m+15=0\\ \Leftrightarrow m=-\dfrac{15}{2}\left(tm\right)\)

\(x^2+5x+m-2\left(1\right)\)

PT (1) là PT bậc 2 có: \(\Delta=5^2-4.\left(m-2\right)=33-4m\)

Để PT có 2 nghiệm phân biệt \(x_1,x_2\) thì \(\Delta>0\Leftrightarrow33-4m>0\Leftrightarrow m< \dfrac{33}{4}\)

Theo định lý Viet ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{1}=-5\\x_1.x_2=\dfrac{c}{a}=\dfrac{m-2}{1}=m-2\end{matrix}\right.\)

Ta có: \(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}=2\Leftrightarrow\dfrac{x_2-1+x_1-1}{\left(x_1-1\right)\left(x_2-1\right)}=2\)

\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)-2}{x_1.x_2-\left(x_1+x_2\right)+1}=2\Leftrightarrow\dfrac{-5-2}{m-2-\left(-5\right)+1}=2\)

\(\Leftrightarrow\dfrac{-7}{m+4}=2\Leftrightarrow m+4=-\dfrac{7}{2}\Leftrightarrow m=-\dfrac{15}{2}\)

\(\Delta=25-4\left(m-2\right)=33-4m>0\Rightarrow m< \dfrac{33}{4}\)

Để biểu thức đề bài xác định \(\Rightarrow\) pt có nghiệm khác 1

\(\Rightarrow1+5+m-2\ne0\Rightarrow m\ne-5\)

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

\(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}=2\Leftrightarrow\dfrac{x_1+x_2-2}{\left(x_1-1\right)\left(x_2-1\right)}=2\)

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

\(\Rightarrow x_1+x_2-2=2x_1x_2-2\left(x_1+x_2\right)+2\)

\(\Leftrightarrow-5-2=2\left(m-2\right)-2.\left(-5\right)+2\)

\(\Leftrightarrow m=-\dfrac{15}{2}\) (thỏa mãn)

\(\Delta'=m^2>0\Rightarrow m\ne0\)

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

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

\(\Leftrightarrow2x_1-\left(-m^2+1\right)+x_2=2\)

\(\Leftrightarrow2x_1+x_2=-m^2+3\)

Kết hợp \(x_1+x_2=2\Rightarrow\left\{{}\begin{matrix}2x_1+x_2=-m^2+3\\x_1+x_2=2\end{matrix}\right.\)

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

Thế vào \(x_1x_2=-m^2+1\)

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

\(\Rightarrow\left[{}\begin{matrix}-m^2+1=0\\m^2+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\pm1\\m=0\left(loại\right)\end{matrix}\right.\)