Đáp án:
Câu 2:
f) A=-2
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a)x\left( {x - y} \right) + \left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + 1} \right)\\
b)\left( {{x^2} - 4x + 4} \right) - {y^2}\\
= {\left( {x - 2} \right)^2} - {y^2}\\
= \left( {x - 2 - y} \right)\left( {x - 2 + y} \right)\\
C2:\\
d)DK:x \ne \pm 3\\
e)A = \dfrac{{3\left( {x - 3} \right) + x + 3 + 18}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4x + 12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{4}{{x - 3}}\\
f)Thay:x = 1\\
A = \dfrac{4}{{1 - 3}} = - 2
\end{array}\)
