Đáp án:
$\begin{array}{l}
3)Do:\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
a)\dfrac{{a + b}}{a} = \dfrac{{bk + b}}{{b.k}} = \dfrac{{b.\left( {k + 1} \right)}}{{b.k}} = \dfrac{{k + 1}}{k}\\
\dfrac{{c + d}}{d} = \dfrac{{d.k + d}}{d} = k + 1 \ne \dfrac{{k + 1}}{k}\\
\Leftrightarrow \dfrac{{a + b}}{a} \ne \dfrac{{c + d}}{d}\\
b)\dfrac{{a - b}}{a} = \dfrac{{bk - b}}{{bk}} = \dfrac{{k - 1}}{k}\\
\dfrac{{c - d}}{d} = \dfrac{{dk - d}}{d} = k - 1 \ne \dfrac{{k - 1}}{k}\\
\Leftrightarrow \dfrac{{a - b}}{a} \ne \dfrac{{c - d}}{d}
\end{array}$
=> Sai đề bài
$\begin{array}{l}
5)\\
a)A + B\\
= {x^2} - 2x + 3x{y^2} - {x^2}y + {x^2}{y^2}\\
+ \left( { - 2{x^2} + 3x{y^2} - 5x} \right)\\
= - {x^2} - 7x + 6x{y^2} - {x^2}y + {x^2}{y^2}\\
b)\\
A + C\\
= {x^2} - 2x + 3x{y^2} - {x^2}y + {x^2}{y^2}\\
+ 3{x^2} - 2x{y^2} - 6\\
= 4{x^2} - 2x + x{y^2} - {x^2}y + {x^2}{y^2} - 6\\
c)\\
A - B\\
= {x^2} - 2x + 3x{y^2} - {x^2}y + {x^2}{y^2}\\
- \left( { - 2{x^2} + 3x{y^2} - 5x} \right)\\
= 3{x^2} + 3x - {x^2}y + {x^2}{y^2}\\
d)B - C\\
= \left( { - 2{x^2} + 3x{y^2} - 5x} \right) - \left( {3{x^2} - 2x{y^2} - 6} \right)\\
= - 5{x^2} + 5x{y^2} - 5x + 6
\end{array}$
