a. \(x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b.\(x^2-\left(3-\sqrt{5}\right)x+2-\sqrt{5}=0\)
\(\Delta=\left[-\left(3-\sqrt{5}\right)\right]^2-4\left(2-\sqrt{5}\right)\)
\(=\left(9-6\sqrt{5}+5\right)-8+4\sqrt{5}\)
\(=6-2\sqrt{5}\)
\(=\left(\sqrt{5}-1\right)^2>0\)
=> pt có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x=\dfrac{3-\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}{2}=\dfrac{2}{2}=1\\x=\dfrac{3-\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}{2}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\end{matrix}\right.\)
c.\(\dfrac{x}{x+2}+\dfrac{x-1}{x}=1\)
\(ĐK:x\ne0;-2\)
\(\Leftrightarrow\dfrac{x^2+\left(x-1\right)\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x\left(x+2\right)}{x\left(x+2\right)}\)
\(\Leftrightarrow x^2+\left(x-1\right)\left(x+2\right)=x\left(x+2\right)\)
\(\Leftrightarrow x^2+x^2+2x-x-2=x^2+2x\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) ( vi-ét )
