a) ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
Vậy \(x\ne-1\) và \(x\ne3\) thì \(A\) xác định
b) \(A=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Leftrightarrow\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Leftrightarrow\dfrac{3x}{2x-6}=0\)
\(\Leftrightarrow3x=0\)
\(\Leftrightarrow x=0\)
Vậy \(x=0\) thì \(A=0\)
a,Để A xác định
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
b,\(A=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)
Để \(A=0\)
\(\Leftrightarrow\dfrac{3x}{2x-6}=0\)
\(\Leftrightarrow3x=0\)
\(\Leftrightarrow x=0\) ( t/m ĐKXĐ )
Vậy x = 0 thì A = 0
