a) Thay m = 0 vào pt ta có:
\(x^2-2(0+1)x+0^2+0-1=0\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow x^2-2x+1-2=0\)
\(\Leftrightarrow\left(x-1\right)^2-2=0\)
\(\Leftrightarrow\left(x-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{matrix}\right.\)
Vậy ....
b) Để phương trình có 2 nghiệm phân biệt thì:
\(\Delta'>0\Leftrightarrow\left(m+1\right)^2-\left(m^2+m-1\right)>0\)
\(\Leftrightarrow m^2+2m+1-m^2-m+1>0\)
\(\Leftrightarrow m+2>0\Leftrightarrow m>-2\)
Theo viet:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1.x_2=m^2+m-1\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=4\)
\(\Leftrightarrow\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=4\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=4\)
\(\Leftrightarrow\dfrac{2m+2}{m^2+m-1}=4\)(đk \(m^2+m-1\ne0\))
\(\Leftrightarrow\dfrac{2m+2}{m^2+m-1}-\dfrac{4m^2+4m-4}{m^2+m-1}=0\) (đk \(m^2+m-1\ne0\))
\(\Leftrightarrow\dfrac{-4m^2-2m+6}{m^2+m-1}=0\) (đk \(m^2+m-1\ne0\))
\(\Leftrightarrow-4m^2-2m+6=0\Leftrightarrow\left[{}\begin{matrix}m=1\left(tm\right)\\m=-\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy
