Đáp án:
\(\begin{array}{l}
1,\\
a,\\
\left( {x - y - 3} \right)\left( {x + y - 3} \right)\\
b,\\
\left( {x - y} \right).\left( {8 - x + y} \right)\\
c,\\
\left( {{x^2} - 3\sqrt 2 x + 9} \right)\left( {{x^2} + 3\sqrt 2 x + 9} \right)\\
2,\\
a,\\
\left[ \begin{array}{l}
x = - 3\\
x = - \dfrac{7}{3}
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = - 1\\
x = 0
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
{x^2} - {y^2} - 6x + 9\\
= \left( {{x^2} - 6x + 9} \right) - {y^2}\\
= \left( {{x^2} - 2.x.3 + {3^2}} \right) - {y^2}\\
= {\left( {x - 3} \right)^2} - {y^3}\\
= \left[ {\left( {x - 3} \right) - y} \right].\left[ {\left( {x - 3} \right) + y} \right]\\
= \left( {x - y - 3} \right)\left( {x + y - 3} \right)\\
b,\\
8x - 8y - {x^2} + 2xy - {y^2}\\
= \left( {8x - 8y} \right) - \left( {{x^2} - 2xy + {y^2}} \right)\\
= 8\left( {x - y} \right) - {\left( {x - y} \right)^2}\\
= \left( {x - y} \right).\left[ {8 - \left( {x - y} \right)} \right]\\
= \left( {x - y} \right).\left( {8 - x + y} \right)\\
c,\\
{x^4} + 81\\
= \left( {{x^4} + 18{x^2} + 81} \right) - 18{x^2}\\
= \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.9 + {9^2}} \right] - {\left( {3\sqrt 2 } \right)^2}{x^2}\\
= {\left( {{x^2} + 9} \right)^2} - {\left( {3\sqrt 2 x} \right)^2}\\
= \left[ {\left( {{x^2} + 9} \right) - 3\sqrt 2 x} \right].\left[ {\left( {{x^2} + 9} \right) + 3\sqrt 2 x} \right]\\
= \left( {{x^2} - 3\sqrt 2 x + 9} \right)\left( {{x^2} + 3\sqrt 2 x + 9} \right)\\
2,\\
a,\\
{\left( {2x + 5} \right)^2} - {\left( {x + 2} \right)^2} = 0\\
\Leftrightarrow \left[ {\left( {2x + 5} \right) - \left( {x + 2} \right)} \right].\left[ {\left( {2x + 5} \right) + \left( {x + 2} \right)} \right] = 0\\
\Leftrightarrow \left( {2x + 5 - x - 2} \right).\left( {2x + 5 + x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right).\left( {3x + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
3x + 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = - \dfrac{7}{3}
\end{array} \right.\\
b,\\
\left( {x + 1} \right)\left( {7x + 2} \right) - 2{x^2} - 4x - 2 = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {7x + 2} \right) - 2.\left( {{x^2} + 2x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {7x + 2} \right) - 2{\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow \left( {x + 1} \right).\left[ {\left( {7x + 2} \right) - 2.\left( {x + 1} \right)} \right] = 0\\
\Leftrightarrow \left( {x + 1} \right).\left( {7x + 2 - 2x - 2} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right).5x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 0
\end{array} \right.
\end{array}\)
