Bài 4:
(1): \(x^2-x-5=0\)
a=1;b=-1;c=-5
Vì a*c=-5<0
nên (1) có hai nghiệm phân biệt trái dấu
Theo Vi-et, ta được: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-1\right)}{1}=1\\x_1\cdot x_2=\dfrac{c}{a}=-5\end{matrix}\right.\)
\(\left(\left|x_1\right|-\left|x_2\right|\right)^2=x_1^2+x_2^2-2\left|x_1\cdot x_2\right|\)
\(=\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|\)
\(=1^2-2\cdot\left(-5\right)-2\left|-5\right|=1+10-10=1\)
=>\(\left[{}\begin{matrix}\left|x_1\right|-\left|x_2\right|=1\\\left|x_1\right|-\left|x_2\right|=-1\end{matrix}\right.\)
TH1: \(\left|x_1\right|-\left|x_2\right|=1\)
=>\(\dfrac{\left|x_1\right|-\left|x_2\right|}{x_1+x_2}=\dfrac{1}{1}=1;\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-5}=-\dfrac{1}{5}\)
\(\dfrac{\left|x_1\right|-\left|x_2\right|}{x_1+x_2}\cdot\dfrac{x_1+x_2}{x_1x_2}=1\cdot\dfrac{-1}{5}=-\dfrac{1}{5}\)
\(\dfrac{\left|x_1\right|-\left|x_2\right|}{x_1+x_2}+\dfrac{x_1+x_2}{x_1x_2}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
Phương trình bậc hai lập được sẽ là: \(a^2-\dfrac{4}{5}a-\dfrac{1}{5}=0\)
TH2: \(\left|x_1\right|-\left|x_2\right|=-1\)
=>\(\dfrac{\left|x_1\right|-\left|x_2\right|}{x_1+x_2}=\dfrac{-1}{1}=-1;\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-5}=-\dfrac{1}{5}\)
\(\dfrac{\left|x_1\right|-\left|x_2\right|}{x_1+x_2}\cdot\dfrac{x_1+x_2}{x_1x_2}=-1\cdot\dfrac{-1}{5}=\dfrac{1}{5}\)
\(\dfrac{\left|x_1\right|-\left|x_2\right|}{x_1+x_2}+\dfrac{x_1+x_2}{x_1x_2}=-1+\dfrac{1}{5}=-\dfrac{4}{5}\)
Phương trình bậc hai lập được sẽ là: \(a^2+\dfrac{4}{5}a+\dfrac{1}{5}=0\)
