Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
2\left( {x + 3} \right)\left( {x + 2} \right) - 4.\left( {x - 2} \right)\left( {x - 5} \right)\\
= 2.\left( {{x^2} + 2x + 3x + 6} \right) - 4.\left( {{x^2} - 5x - 2x + 10} \right)\\
= 2.\left( {{x^2} + 5x + 6} \right) - 4\left( {{x^2} - 7x + 10} \right)\\
= 2{x^2} + 10x + 12 - 4{x^2} + 28x - 40\\
= - 2{x^2} + 38x - 28\\
b,\\
2x\left( {x - 3} \right) - {\left( {x - 1} \right)^2} - \left( {x + 2} \right)\left( {x - 2} \right) - 5\\
= \left( {2{x^2} - 6x} \right) - \left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - {2^2}} \right) - 5\\
= 2{x^2} - 6x - {x^2} + 2x - 1 - {x^2} + 4 - 5\\
= - 4x - 2\\
2,\\
{x^2}\left( {y - 2} \right) + 2x\left( {2 - y} \right) + \left( {y - 2} \right)\\
= {x^2}\left( {y - 2} \right) - 2x.\left( {y - 2} \right) + \left( {y - 2} \right)\\
= \left( {y - 2} \right)\left( {{x^2} - 2x + 1} \right)\\
= \left( {y - 2} \right){\left( {x - 1} \right)^2}
\end{array}\)
Em xem lại đề câu 3 nhé!
