1: Thay m=2 vào phương trình (1), ta được:
\(x^2-2x-1=0\)
=>\(x^2-2x+1-2=0\)
=>\(\left(x-1\right)^2-2=0\)
=>\(\left(x-1\right)^2=2\)
=>\(\left[{}\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{matrix}\right.\)
2: \(x^2-mx-1=0\)
\(\text{Δ}=\left(-m\right)^2-4\cdot1\cdot\left(-1\right)=m^2+4>0\)
=>Phương trình luôn có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{m-\sqrt{m^2+4}}{2}\\x_2=\dfrac{m+\sqrt{m^2+4}}{2}\end{matrix}\right.\)
\(\left|x_1\right|-\left|x_2\right|=6\)
=>\(\left|\dfrac{m-\sqrt{m^2+4}}{2}\right|-\left|\dfrac{m+\sqrt{m^2+4}}{2}\right|=6\)
=>\(\left|m-\sqrt{m^2+4}\right|-\left|m+\sqrt{m^2+4}\right|=12\)
=>\(\left|\dfrac{m^2-m^2-4}{m+\sqrt{m^2+4}}\right|-\left|m+\sqrt{m^2+4}\right|=12\)
=>\(\dfrac{4}{\left|m+\sqrt{m^2+4}\right|}-\left|m+\sqrt{m^2+4}\right|=12\)(2)
Đặt \(\left|m+\sqrt{m^2+4}\right|=a\left(a>0\right)\)
Phương trình (2) sẽ trở thành \(\dfrac{4}{a}-a=12\)
=>\(\dfrac{4-a^2}{a}=12\)
=>\(4-a^2=12a\)
=>\(a^2+12a-4=0\)
=>\(\left(a+6\right)^2-40=0\)
=>\(\left(a+6\right)^2=40\)
=>\(\left[{}\begin{matrix}a+6=2\sqrt{10}\\a+6=-2\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=2\sqrt{10}-6\left(nhận\right)\\a=-2\sqrt{10}-6\left(loại\right)\end{matrix}\right.\)
=>\(\left|m+\sqrt{m^2+4}\right|=2\sqrt{10}-6\)
=>\(\left[{}\begin{matrix}m+\sqrt{m^2+4}=2\sqrt{10}-6\\m+\sqrt{m^2+4}=-2\sqrt{10}+6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\sqrt{m^2+4}=-m+2\sqrt{10}-6\left(3\right)\\\sqrt{m^2+4}=-m-2\sqrt{10}+6\left(4\right)\end{matrix}\right.\)
(3): \(\sqrt{m^2+4}=-m+2\sqrt{10}-6\)
=>\(\left\{{}\begin{matrix}-m+2\sqrt{10}-6>0\\m^2+4=\left(-m+2\sqrt{10}-6\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-m>-2\sqrt{10}+6\\m^2+4=m^2-2m\left(2\sqrt{10}-6\right)+\left(2\sqrt{10}-6\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 2\sqrt{10}-6\\-2m\left(2\sqrt{10}-6\right)+76-24\sqrt{10}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 2\sqrt{10}-6\\m\left(-4\sqrt{10}+12\right)=4-76+24\sqrt{10}=24\sqrt{10}-72\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 2\sqrt{10}-6\\m=-6\end{matrix}\right.\)
=>m=-6(nhận)
(4): \(\sqrt{m^2+4}=-m-2\sqrt{10}+6\)
=>\(\left\{{}\begin{matrix}-m-2\sqrt{10}+6>0\\m^2+4=\left(-m-2\sqrt{10}+6\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-m>2\sqrt{10}-6\\m^2+4=\left(m+2\sqrt{10}-6\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -2\sqrt{10}+6\\m^2+4=m^2+2m\left(2\sqrt{10}-6\right)+76-24\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -2\sqrt{10}+6\\m\left(4\sqrt{10}-12\right)+76-24\sqrt{10}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -2\sqrt{10}+6\\m\left(4\sqrt{10}-12\right)=-72+24\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -2\sqrt{10}+6\\m=6\end{matrix}\right.\)
=>\(m\in\varnothing\)
