Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\left( {x + 6} \right)\left( {x - 1} \right)\\
b,\,\,\,\,\left( {x + 3} \right)\left( {2x - 1} \right)\\
c,\,\,\,\,\left( {x - 1} \right)\left( {2x + 5} \right)\\
d,\,\,\,\,\left( {x + 3} \right)\left( {x - 2} \right)\\
g,\,\,\,\,\left( {3x - 2} \right)\left( { - 2x + 1} \right)\\
h,\,\,\,\,\left( {x - 3} \right)\left( { - 5x + 1} \right)\\
i,\,\,\,\,\left( {x + 3} \right)\left( {x + 1} \right)\\
k,\,\,\,\,\left( {x + 4} \right)\left( {x - 3} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
{x^2} + 5x - 6 = \left( {{x^2} + 6x} \right) + \left( { - x - 6} \right)\\
= x\left( {x + 6} \right) - \left( {x + 6} \right) = \left( {x + 6} \right)\left( {x - 1} \right)\\
b,\\
2{x^2} + 5x - 3 = \left( {2{x^2} + 6x} \right) + \left( { - x - 3} \right)\\
= 2x.\left( {x + 3} \right) - \left( {x + 3} \right) = \left( {x + 3} \right)\left( {2x - 1} \right)\\
c,\\
2{x^2} + 3x - 5 = \left( {2{x^2} - 2x} \right) + \left( {5x - 5} \right)\\
= 2x.\left( {x - 1} \right) + 5\left( {x - 1} \right) = \left( {x - 1} \right)\left( {2x + 5} \right)\\
d,\\
{x^2} + x - 6 = \left( {{x^2} + 3x} \right) + \left( { - 2x - 6} \right)\\
= x.\left( {x + 3} \right) - 2.\left( {x + 3} \right) = \left( {x + 3} \right)\left( {x - 2} \right)\\
g,\\
7x - 6{x^2} - 2 = - 6{x^2} + 7x - 2\\
= \left( { - 6{x^2} + 4x} \right) + \left( {3x - 2} \right)\\
= - 2x.\left( {3x - 2} \right) + \left( {3x - 2} \right)\\
= \left( {3x - 2} \right)\left( { - 2x + 1} \right)\\
h,\\
16x - 5{x^2} - 3 = - 5{x^2} + 16x - 3\\
= \left( { - 5{x^2} + 15x} \right) + \left( {x - 3} \right)\\
= - 5x.\left( {x - 3} \right) + \left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( { - 5x + 1} \right)\\
i,\\
{x^2} + 4x + 3 = \left( {{x^2} + 3x} \right) + \left( {x + 3} \right)\\
= x\left( {x + 3} \right) + \left( {x + 3} \right)\\
= \left( {x + 3} \right)\left( {x + 1} \right)\\
k,\\
{x^2} + x - 12 = \left( {{x^2} + 4x} \right) + \left( { - 3x - 12} \right)\\
= x\left( {x + 4} \right) - 3.\left( {x + 4} \right) = \left( {x + 4} \right)\left( {x - 3} \right)
\end{array}\)
Em xem lại đề câu e nhé!
