Đáp án:
\(\begin{array}{l}
1,\\
a,\,\,\,\,56x + 32\\
b,\,\,\,\, - 4x + 5\\
c,\,\,\,\,2\\
2,\\
a,\,\,\,\,3xy\left( {1 - 2y} \right)\\
b,\,\,\,\,3x{\left( {x + 1} \right)^2}\\
c,\,\,\,\,\,\left( {x - y + 2} \right)\left( {x + y + 2} \right)\\
d,\,\,\,\,\left( {x + 1} \right)\left( {{x^2} - 2x + 2} \right)\\
3,\\
a,\,\,\,\left[ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.\\
b,\,\,\,\,\left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
c,\,\,\,\,x = - 1\\
d,\,\,\,\,\left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
{\left( {7x + 4} \right)^2} - \left( {7x + 4} \right)\left( {7x - 4} \right)\\
= \left( {7x + 4} \right).\left[ {\left( {7x + 4} \right) - \left( {7x - 4} \right)} \right]\\
= \left( {7x + 4} \right).\left( {7x + 4 - 7x + 4} \right)\\
= \left( {7x + 4} \right).8\\
= 56x + 32\\
b,\\
8x\left( {x - 2} \right) - 3.\left( {{x^2} - 4x - 5} \right) - 5{x^2}\\
= \left( {8{x^2} - 16x} \right) - \left( {3{x^2} - 12x - 5} \right) - 5{x^2}\\
= 8{x^2} - 16x - 3{x^2} + 12x + 5 - 5{x^2}\\
= \left( {8{x^2} - 3{x^2} - 5{x^2}} \right) + \left( { - 16x + 12x} \right) + 5\\
= - 4x + 5\\
c,\\
{\left( {x + 1} \right)^3} - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - 3x\left( {x + 1} \right)\\
= \left( {{x^3} + 3.{x^2}.1 + 3.x{{.1}^2} + {1^3}} \right) - \left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right) - \left( {3{x^2} + 3x} \right)\\
= \left( {{x^3} + 3{x^2} + 3x + 1} \right) - \left( {{x^3} - {1^3}} \right) - \left( {3{x^2} + 3x} \right)\\
= {x^3} + 3{x^2} + 3x + 1 - {x^3} + 1 - 3{x^2} - 3x\\
= \left( {{x^3} - {x^3}} \right) + \left( {3{x^2} - 3{x^2}} \right) + \left( {3x - 3x} \right) + \left( {1 + 1} \right)\\
= 2\\
2,\\
a,\\
3xy - 6x{y^2} = 3xy - 3xy.2y = 3xy\left( {1 - 2y} \right)\\
b,\\
3{x^3} + 6{x^2} + 3x = 3x.\left( {{x^2} + 2x + 1} \right) = 3x.\left( {{x^2} + 2.x.1 + {1^2}} \right) = 3x{\left( {x + 1} \right)^2}\\
c,\\
{x^2} + 4x + 4 - {y^2} = \left( {{x^2} + 4x + 4} \right) - {y^2}\\
= \left( {{x^2} + 2.x.2 + {2^2}} \right) - {y^2} = {\left( {x + 2} \right)^2} - {y^2}\\
= \left[ {\left( {x + 2} \right) - y} \right].\left[ {\left( {x + 2} \right) + y} \right]\\
= \left( {x - y + 2} \right)\left( {x + y + 2} \right)\\
d,\\
{x^3} - {x^2} + 2 = \left( {{x^3} + {x^2}} \right) + \left( { - 2{x^2} - 2x} \right) + \left( {2x + 2} \right)\\
= {x^2}\left( {x + 1} \right) - 2x.\left( {x + 1} \right) + 2.\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - 2x + 2} \right)\\
3,\\
a,\\
{x^2} - 16 = 0\\
\Leftrightarrow {x^2} - {4^2} = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.\\
b,\\
2\left( {x + 3} \right) - {x^2} - 3x = 0\\
\Leftrightarrow 2\left( {x + 3} \right) - \left( {{x^2} + 3x} \right) = 0\\
\Leftrightarrow 2.\left( {x + 3} \right) - x\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
c,\\
{x^3} + {x^2} + x + 1 = 0\\
\Leftrightarrow \left( {{x^3} + {x^2}} \right) + \left( {x + 1} \right) = 0\\
\Leftrightarrow {x^2}\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 1} \right) = 0\\
{x^2} + 1 \ge 1 > 0,\,\,\,\forall x\\
\Rightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\\
d,\\
{x^2} - x = 6\\
\Leftrightarrow {x^2} - x - 6 = 0\\
\Leftrightarrow \left( {{x^2} - 3x} \right) + \left( {2x - 6} \right) = 0\\
\Leftrightarrow x\left( {x - 3} \right) + 2.\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.
\end{array}\)
