3. For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the range of $x$ that makes the inequality $x^{2}+p x>4 x+p-3$ hold is $\qquad$ .
$$
\begin{array}{l}
\text { 3. } x3 \text {. } \\
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
\end{array}
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right.
$$
$$
\text { 3. } x3 \text {. }
$$
$$
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l } 
{ x > 3 - p } \\
{ x