Bài 1.
a) \(10{x^2} + 10xy + 5x + 5y \)
\(= \left( {10{x^2} + 10xy} \right) + \left( {5x + 5y} \right)\)
\( = 10x\left( {x + y} \right) + 5\left( {x + y} \right) \)
\(= \left( {x + y} \right)\left( {10x + 5} \right) \)
\(= 5\left( {x + y} \right)\left( {2x + 1} \right).\)
b) \(5ay - 3bx + ax - 5by \)
\(= \left( {5ay + ax} \right) + \left( { - 3bx - 15by} \right)\)
\( = a\left( {5y + x} \right) - 3b\left( {x + 5y} \right) \)
\(= \left( {5y + x} \right)\left( {a - 3b} \right).\)
c) \({x^3} + {x^2} - x - 1 \)
\(= \left( {{x^3} + {x^2}} \right) - \left( {x + 1} \right)\)
\(= {x^2}\left( {x + 1} \right) - \left( {x + 1} \right)\)
\( = \left( {x + 1} \right)\left( {{x^2} - 1} \right) \)
\(= {\left( {x + 1} \right)^2}\left( {x - 1} \right).\)
Bài 2.
a) \(x\left( {x - 2} \right) + x - 2 = \left( {x - 2} \right)\left( {x + 1} \right)\)
Vậy \(\left( {x - 2} \right)\left( {x + 1} \right) = 0\)
\(\Rightarrow x - 2 = 0\) hoặc \(x + 1 = 0\)
\( \Rightarrow x = 2\) hoặc \(x = - 1.\)
b) \({x^3} + {x^2} + x + 1 \)
\(= {x^2}\left( {x + 1} \right) + \left( {x + 1} \right) \)
\(= \left( {x + 1} \right)\left( {{x^2} + 1} \right)\)
Vậy \(\left( {x + 1} \right)\left( {{x^2} + 1} \right) = 0 \Rightarrow x + 1 = 0\)
\( \Rightarrow x = - 1\) (vì \({x^2} + 1 > 0,\) với mọi x).
