Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}a){x^2} + x - {y^2} + y\\ = \left( {{x^2} - {y^2}} \right) + \left( {x + y} \right)\\ = \left( {x + y} \right)\left( {x - y} \right) + \left( {x + y} \right)\\ = \left( {x + y} \right)\left( {x - y + 1} \right)\end{array}\)
\(\begin{array}{l}b)3{x^2} + 3{y^2} - 6xy - 12\\ = 3\left( {{x^2} + {y^2} - 2xy - 4} \right)\\ = 3\left[ {{{\left( {x - y} \right)}^2} - {2^2}} \right]\\ = 3\left( {x - y - 2} \right)\left( {x - y + 2} \right)\end{array}\)
\(\begin{array}{l}c)3x + 3y - {x^2} - 2xy - {y^2}\\ = 3\left( {x + y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\ = 3\left( {x + y} \right) - {\left( {x + y} \right)^2}\\ = \left( {x + y} \right)\left( {3 - x - y} \right)\end{array}\)
\(\begin{array}{l}d){x^3} - x + 3{x^2}y + 3x{y^2} - y + {y^3}\\ = \left( {{x^3} + {y^3}} \right) - \left( {x + y} \right) + 3xy\left( {x + y} \right)\\ = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - \left( {x + y} \right) + 3xy\left( {x + y} \right)\\ = \left( {x + y} \right)\left[ {{x^2} - xy + {y^2} - 1 + 3xy} \right]\\ = \left( {x + y} \right)\left( {{x^2} + 2xy + {y^2} - 1} \right)\\ = \left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} - 1} \right]\end{array}\)
