a: \(x^2-9=0\)
=>\(\left(x-3\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
b: \(x^2-6x+9=0\)
=>\(x^2-2\cdot x\cdot3+3^2=0\)
=>\(\left(x-3\right)^2=0\)
=>x-3=0
=>x=3
c: \(x^2-5x+6=0\)
=>\(x^2-2x-3x+6=0\)
=>\(x\cdot\left(x-2\right)-3\left(x-2\right)=0\)
=>(x-2)(x-3)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
d: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(\dfrac{3}{x-1}+\dfrac{4}{x+1}=\dfrac{5}{x^2-1}\)
=>\(\dfrac{3}{x-1}+\dfrac{4}{x+1}=\dfrac{5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{3\left(x+1\right)+4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{\left(x-1\right)\left(x+1\right)}\)
=>3x+3+4x-4=5
=>7x-1=5
=>7x=6
=>\(x=\dfrac{6}{7}\left(nhận\right)\)
