Ta có :
\(x^2-mx-1=9\)
\(\Leftrightarrow x^2-mx-10=0\)
\(\Delta=m^2+40>0\)
Nên pt có 2 nghiệm phân biệt theo \(x_1,x_2\)
Theo hệ thức vi-ét ta có :
\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=-10\end{matrix}\right.\)
Ta lại có :
\(\left(x_1^2-1\right)\left(x_2^2-1\right)=-1\)
\(\Leftrightarrow x_1^2x_2^2-x_1^2-x_2^2+2=0\)
\(\Leftrightarrow\left(x_1x_2\right)^2-\left[\left(x_1+x_2\right)^2-2.x_1x_2\right]+2=0\)
\(\Leftrightarrow100-m^2-20+2=0\)
\(\Leftrightarrow-m^2+82=0\)
\(\Leftrightarrow m=\sqrt{82}\)
Vậy \(m=\sqrt{82}\)
x^2-mx-10=0
c/a<0= > co nghiem moi m
(x1^2-1)(x2^2-1)=(x1-1)(x2-1)(x1+1)(x2+1)
[x1.x2-(x1+x3)+1][x1.x2+(x1+x2)+1]
vi et
<=>(-9-m)(-9+m)=-1
<=>81-m^2=-1
m=±√(82)
