a) \(x^2-mx+m-2=0\) (1)
Thay \(m=-2\) vào pt (1), ta được:
\(x^2-\left(-2\right)\cdot x+\left(-2\right)-2=0\)
\(\Leftrightarrow x^2+2x-4=0\)
\(\Leftrightarrow\left(x+1\right)^2-5=0\)
\(\Leftrightarrow\left(x+1-\sqrt{5}\right)\left(x+1+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1-\sqrt{5}=0\\x+1+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)
Vậy pt (1) có nghiệm là \(x\in\left\{-1+\sqrt{5};-1-\sqrt{5}\right\}\) khi m = -2.
b) \(\Delta=m^2-4\left(m-2\right)=m^2-4m+8=\left(m-2\right)^2+4>0;\forall m\)
\(\Rightarrow\) Pt (1) có 2 nghiệm phân biệt với mọi m
c) Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)
Theo đề ra, ta có: \(\dfrac{x_1^2-2}{x_1-1}\cdot\dfrac{x_2^2-2}{x_2-1}=4\)
\(\Leftrightarrow\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\)
\(\Leftrightarrow\dfrac{x_1^2x_2^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Leftrightarrow\dfrac{\left(x_1x_2\right)^2+4x_1x_2+4-2\left(x_1+x_2\right)^2}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(m-2\right)^2+4\left(m-2\right)+4-2m^2}{m-2-m+1}=4\)
\(\Leftrightarrow\dfrac{\left(m-2+2\right)^2-2m^2}{-1}=4\)
\(\Leftrightarrow m^2-2m^2=-4\)
\(\Leftrightarrow-m^2=-4\)
\(\Leftrightarrow m^2=4\)
\(\Leftrightarrow m=\pm2\)
Vậy: ...
a: Thay m=-2 vào (1), ta được:
\(x^2-\left(-2\right)x+\left(-2\right)-2=0\)
=>\(x^2+2x-4=0\)
=>\(x^2+2x+1-5=0\)
=>\(\left(x+1\right)^2=5\)
=>\(x+1=\pm\sqrt{5}\)
=>\(x=-1\pm\sqrt{5}\)
b: \(\text{Δ}=\left(-m\right)^2-4\cdot1\cdot\left(m-2\right)\)
\(=m^2-4m+8=m^2-4m+4+4=\left(m-2\right)^2+4>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
