Đáp án:
\(\begin{array}{l}
3,\\
a,\,\,\,A = {a^2}.{\left( {a - 6} \right)^2}\\
b,\,\,\,B = \left( {x - 2y + 3} \right)\left( {x + 2y + 3} \right)\\
c,\,\,\,C = \left( {a + 3} \right)\left( {2a + 1} \right)\\
4,\\
a,\,\,\left[ \begin{array}{l}
x = - 2016\\
x = 15
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = 0\\
x = 10
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
a,\,\,\,A = {a^4} - 12{a^3} + 36{a^2}\\
= {a^2}.\left( {{a^2} - 12a + 36} \right)\\
= {a^2}.\left( {{a^2} - 2.a.6 + {6^2}} \right)\\
= {a^2}.{\left( {a - 6} \right)^2}\\
b,\,\,\,B = {x^2} - 4{y^2} + 6x + 9\\
= \left( {{x^2} + 6x + 9} \right) - 4{y^2}\\
= \left( {{x^2} + 2.x.3 + {3^2}} \right) - {\left( {2y} \right)^2}\\
= {\left( {x + 3} \right)^2} - {\left( {2y} \right)^2}\\
= \left[ {\left( {x + 3} \right) - 2y} \right].\left[ {\left( {x + 3} \right) + 2y} \right]\\
= \left( {x - 2y + 3} \right)\left( {x + 2y + 3} \right)\\
c,\,\,\,C = 2{a^2} + 7a + 3\\
= \left( {2{a^2} + 6a} \right) + \left( {a + 3} \right)\\
= 2a.\left( {a + 3} \right) + \left( {a + 3} \right)\\
= \left( {a + 3} \right)\left( {2a + 1} \right)\\
4,\\
a,\,\,15\left( {x + 2016} \right) - {x^2} - 2016x = 0\\
\Leftrightarrow 15.\left( {x + 2016} \right) - \left( {{x^2} + 2016x} \right) = 0\\
\Leftrightarrow 15.\left( {x + 2016} \right) - x.\left( {x + 2016} \right) = 0\\
\Leftrightarrow \left( {x + 2016} \right).\left( {15 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2016 = 0\\
15 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 2016\\
x = 15
\end{array} \right.\\
b,\,\,\,\left( {{x^2} + 2} \right)\left( {x - 5} \right) - {\left( {x - 2} \right)^3} = - 2\\
\Leftrightarrow \left( {{x^2} + 2} \right).x - \left( {{x^2} + 2} \right).5 - \left( {{x^3} - 3.{x^2}.2 + 3.x{{.2}^2} - {2^3}} \right) = 2\\
\Leftrightarrow \left( {{x^3} + 2x} \right) - \left( {5{x^2} + 10} \right) - \left( {{x^3} - 6{x^2} + 12x - 8} \right) = - 2\\
\Leftrightarrow {x^3} + 2x - 5{x^2} - 10 - {x^3} + 6{x^2} - 12x + 8 + 2 = 0\\
\Leftrightarrow \left( {{x^3} - {x^3}} \right) + \left( { - 5{x^2} + 6{x^2}} \right) + \left( {2x - 12x} \right) + \left( { - 10 + 8 + 2} \right) = 0\\
\Leftrightarrow {x^2} - 10x = 0\\
\Leftrightarrow x\left( {x - 10} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 10 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 10
\end{array} \right.
\end{array}\)
