Bài 10:
a: Thay m=0 vào phương trình, ta được:
\(x^2-\left(0-2\right)x-6=0\)
=>\(x^2+2x-6=0\)
=>\(x^2+2x+1-7=0\)
=>\(\left(x+1\right)^2=7\)
=>\(x+1=\pm\sqrt{7}\)
=>\(x=-1\pm\sqrt{7}\)
b: \(x^2-\left(m-2\right)x-6=0\)
a=1; b=-(m-2); c=-6
Vì \(a\cdot c=1\cdot\left(-6\right)=-6< 0\)
nên phương trình luôn có hai nghiệm phân biệt
c: Theo Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left[-\left(m-2\right)\right]}{1}=m-2\\x_1x_2=\dfrac{-6}{1}=-6\end{matrix}\right.\)
\(x_2^2-x_1x_2+\left(m-2\right)x=16\)
=>\(x_2^2-x_1x_2+\left(x_1+x_2\right)x_1=16\)
=>\(x_1^2-x_1x_2+x_1x_2+x_2^2=16\)
=>\(x_1^2+x_2^2=16\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=16\)
=>\(\left(m-2\right)^2-2\left(-6\right)=16\)
=>\(\left(m-2\right)^2=4\)
=>\(\left[{}\begin{matrix}m-2=2\\m-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=4\left(nhận\right)\\m=0\left(nhận\right)\end{matrix}\right.\)
Bài 9:
a: \(x^2-2mx+4m-4=0\)
\(\text{Δ}=\left(-2m\right)^2-4\left(4m-4\right)\)
\(=4m^2-16m+16=\left(2m-4\right)^2\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>\(\left(2m-4\right)^2>0\)
=>\(2m-4\ne0\)
=>\(2m\ne4\)
=>\(m\ne2\)
b: theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-2m\right)}{1}=2m\\x_1x_2=\dfrac{c}{a}=4m-4\end{matrix}\right.\)
\(x_1^2+2mx_2-8m+5=0\)
=>\(x_1^2+x_2\left(x_1+x_2\right)-8m+5=0\)
=>\(\left(x_1^2+x_2^2\right)+x_1x_2-8m+5=0\)
=>\(\left(x_1+x_2\right)^2-x_1x_2-8m+5=0\)
=>\(\left(2m\right)^2-4m+4-8m+5=0\)
=>\(4m^2-12m+9=0\)
=>\(\left(2m-3\right)^2=0\)
=>2m-3=0
=>2m=3
=>\(m=\dfrac{3}{2}\)