bài này sử dụng định lí vi-ét nha

\(a,x^2-\left(2m-3\right)x+m^2=0-vô-ngo\)

\(\Leftrightarrow\Delta< 0\Leftrightarrow[-\left(2m-3\right)]^2-4m^2< 0\Leftrightarrow m>\dfrac{3}{4}\)

\(b,\left(m-1\right)x^2-2mx+m-2=0\)

\(m-1=0\Leftrightarrow m=1\Rightarrow-2x-1=0\Leftrightarrow x=-0,5\left(ktm\right)\)

\(m-1\ne0\Leftrightarrow m\ne1\Rightarrow\Delta'< 0\Leftrightarrow\left(-m\right)^2-\left(m-2\right)\left(m-1\right)< 0\Leftrightarrow m< \dfrac{2}{3}\)

\(c,\left(2-m\right)x^2-2\left(m+1\right)x+4-m=0\)

\(2-m=0\Leftrightarrow m=2\Rightarrow-6x+2=0\Leftrightarrow x=\dfrac{1}{3}\left(ktm\right)\)

\(2-m\ne0\Leftrightarrow m\ne2\Rightarrow\Delta'< 0\Leftrightarrow[-\left(m+1\right)]^2-\left(4-m\right)\left(2-m\right)< 0\Leftrightarrow m< \dfrac{7}{8}\)

\(P\left(1\right)=Q\left(-1\right)\)

\(\Leftrightarrow1^2+2m+m^2=\left(-1\right)^2+\left(2m+1\right)\cdot\left(-1\right)+m^2\)

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

=>2m+1=-2m

=>4m=-1

hay m=-1/4

Ta có : 

\(P\left(-1\right)=Q\left(1\right)\)

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

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

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

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

\(\Leftrightarrow\)\(-4m=1\)

\(\Leftrightarrow\)\(m=\frac{1}{-4}\)

\(\Leftrightarrow\)\(m=\frac{-1}{4}\)

Vậy giá trị của \(m\) là \(\frac{-1}{4}\) khi \(P\left(-1\right)=Q\left(1\right)\)

Chúc bạn học tốt ~ 

P(-1) = 1 - 2m + m² = Q(1) = 1 + 2m + 1 + m² 

=> m² - 2m = m² + 2m + 1

=> 4m +1 = 0

=> m =\(\frac{-1}{4}\)

p(1) = 1² + 2m . 1 + m²    (dấu . là nhân nhé)

=> p(1) = 1 + 2m + m²

q(-1) = (-1)² + (2m + 1) . (-1) + m²

=> q(-1) = 1 - (2m + 1) + m² = 1 - 2m - 1 + m² = -2m + m²

p(1) = q(-1)

=> 1 + 2m + m²  =- 2m + m²

=>  1 + 2m + 2m = m² - m²   (chuyển vế)

=> 1 + 4m = 0

=> 4m = -1

=> m = -1/4