Đáp án:
B2:
d) \(\dfrac{{7\left( {x + 1} \right)}}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a,{(x + 2)^2} - (x - 2)(x + 2) = 0\\
\to \left( {x + 2} \right)\left( {x + 2 - x + 2} \right) = 0\\
\to 4\left( {x + 2} \right) = 0\\
\to x = - 2\\
b)x\left( {x - 3} \right) - 5\left( {x - 3} \right) = 0\\
\to \left( {x - 3} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = 5
\end{array} \right.\\
B2:\\
a)\dfrac{{12{x^3}{y^2}}}{{18x{y^5}}} = \dfrac{{2{x^2}}}{{3{y^3}}}\\
b)\dfrac{{12xy}}{{4xy}} = 3\\
c)\dfrac{{3{x^2} - 12x + 12}}{{{x^4} - 8x}}\\
= \dfrac{{3{{\left( {x - 2} \right)}^2}}}{{x\left( {x - 2} \right)\left( {x + 2x + 4} \right)}}\\
= \dfrac{{3\left( {x - 2} \right)}}{{x\left( {x + 2x + 4} \right)}}\\
d)\dfrac{{7{x^2} + 14x + 7}}{{3x + 3}} = \dfrac{{7{{\left( {x + 1} \right)}^2}}}{{3\left( {x + 1} \right)}}\\
= \dfrac{{7\left( {x + 1} \right)}}{3}
\end{array}\)
