Đáp án:
$\begin{array}{l}
1)a)5x\left( {3x - 2} \right)\\
= 15{x^2} - 10x\\
b)\left( {2x + 5} \right)\left( {x - 3} \right)\\
= 2{x^2} - 6x + 5x - 15\\
= 2{x^2} - x - 15\\
c)\dfrac{{4xy + 7}}{{3x}} + \dfrac{{2xy - 7}}{{3x}}\\
= \dfrac{{4xy + 7 + 2xy - 7}}{{3x}}\\
= \dfrac{{6xy}}{{3x}}\\
= 2y\\
d)\dfrac{{{x^2} + 49}}{{{x^2} - 7x}} - \dfrac{{14}}{{x - 7}}\\
= \dfrac{{{x^2} + 49 - 14x}}{{x\left( {x - 7} \right)}}\\
= \dfrac{{{{\left( {x - 7} \right)}^2}}}{{x\left( {x - 7} \right)}}\\
= \dfrac{{x - 7}}{x}\\
2)a)ax - ay\\
= a.\left( {x - y} \right)\\
b)2{x^3} - 50x\\
= 2x\left( {{x^2} - 25} \right)\\
= 2x\left( {x - 5} \right)\left( {x + 5} \right)\\
c){x^2} - xy + 4x - 4y\\
= x\left( {x - y} \right) + 4\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + 4} \right)\\
d){x^2} - 4x + 4 - {y^2}\\
= {\left( {x - 2} \right)^2} - {y^2}\\
= \left( {x - 2 - y} \right)\left( {x - 2 + y} \right)\\
3)a)7x\left( {x - 3} \right) + 4\left( {x - 3} \right) = 0\\
\Rightarrow \left( {x - 3} \right)\left( {7x + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 3 = 0\\
7x + 4 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{4}{7}
\end{array} \right.\\
Vậy\,x = 3;x = - \dfrac{4}{7}\\
b){x^2} - x + \dfrac{1}{4} = 0\\
\Rightarrow 4{x^2} - 4x + 1 = 0\\
\Rightarrow {\left( {2x - 1} \right)^2} = 0\\
\Rightarrow x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{1}{2}
\end{array}$
