Descomposición en factores y simplificación

Las siguientes expresiones podrán descomponerse mediante productos notables o sacando factor común

\(\bullet \dfrac {x^{2}-9} {x^{2}-6x+9}=\dfrac {\left( x+3\right)\enclose{updiagonalstrike} {\left( x-3\right)} } {\left( x-3\right) ^\enclose{updiagonalstrike}{{2}}}=\boxed{\dfrac {x+3} {x-3}}\)

\(\bullet \dfrac {3x^{2}+3} {3x^{2}-3}=\dfrac {\enclose{updiagonalstrike}{3}\enclose{updiagonalstrike}{\left( x+1\right) }} {\enclose{updiagonalstrike}{3}\enclose{updiagonalstrike}{\left( x+1\right)} \left( x-1\right) }=\boxed{\dfrac {1} {x-1}}\)

\(\bullet \dfrac {x^{2}+2x+1} {5x^{2}+5x}=\dfrac {\left( x+1\right) ^\enclose{updiagonalstrike}{{2}}} {5\enclose{updiagonalstrike}{\left( x+1\right)} }=\boxed{\dfrac {x+1} {5x}}\)

\(\bullet \dfrac {5x+15} {x^{2}+6x+9}=\dfrac {5\enclose{updiagonalstrike}{\left( x+3\right)} } {\left( x+3\right) ^\enclose{updiagonalstrike}{{2}}}=\boxed{\dfrac {5} {x+3}}\)

\(\bullet \dfrac {3x^{2}+6x+3} {5x^{2}+5x}=\dfrac {3\left( x+1\right) ^\enclose{updiagonalstrike}{{2}}} {5x\enclose{updiagonalstrike}{\left( x+1\right)} }=\boxed{\dfrac {3\left( x+1\right) } {5x}}\)

Si tenéis alguna duda, dejadme un mensaje muchas gracias