a) x^3-4x^2 – 8x + 8
b) 1+6x-6x^2-x^3
c) x^4 – 4x^2 +4x – 1
d) (xy+1)^2 – (x+y)^2
e) X^2 – 7x +12′
f) x^2 – 5x – 14
g) 4x^2-3x-1
a) x^3-4x^2 – 8x + 8
b) 1+6x-6x^2-x^3
c) x^4 – 4x^2 +4x – 1
d) (xy+1)^2 – (x+y)^2
e) X^2 – 7x +12′
f) x^2 – 5x – 14
g) 4x^2-3x-1
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^3} – 4{x^2} – 8x + 8\\
= \left( {{x^3} + 2{x^2}} \right) – \left( {6{x^2} + 12x} \right) + \left( {4x + 8} \right)\\
= {x^2}\left( {x + 2} \right) – 6x\left( {x + 2} \right) + 4\left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {{x^2} – 6x + 4} \right)\\
b,\\
1 + 6x – 6{x^2} – {x^3}\\
= \left( {1 – {x^3}} \right) + \left( {6x – 6{x^2}} \right)\\
= \left( {1 – x} \right)\left( {1 + x + {x^2}} \right) + 6x\left( {1 – x} \right)\\
= \left( {1 – x} \right).\left( {1 + x + {x^2} + 6x} \right)\\
= \left( {1 – x} \right)\left( {1 + 7x + {x^2}} \right)\\
c,\\
{x^4} – 4{x^2} + 4x – 1\\
= \left( {{x^4} – 1} \right) – \left( {4{x^2} – 4x} \right)\\
= \left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right) – 4x\left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right) – 4x\left( {x – 1} \right)\\
= \left( {x – 1} \right).\left[ {\left( {x + 1} \right)\left( {{x^2} + 1} \right) – 4x} \right]\\
= \left( {x – 1} \right).\left[ {\left( {{x^3} + x + {x^2} + 1} \right) – 4x} \right]\\
= \left( {x – 1} \right).\left( {{x^3} + {x^2} – 3x + 1} \right)\\
= \left( {x – 1} \right).\left[ {\left( {{x^3} – {x^2}} \right) + \left( {2{x^2} – 2x} \right) – \left( {x – 1} \right)} \right]\\
= \left( {x – 1} \right).\left[ {{x^2}\left( {x – 1} \right) + 2x\left( {x – 1} \right) – \left( {x – 1} \right)} \right]\\
= \left( {x – 1} \right).\left( {x – 1} \right).\left( {{x^2} + 2x – 1} \right)\\
= {\left( {x – 1} \right)^2}\left( {{x^2} + 2x – 1} \right)\\
d,\\
{\left( {xy + 1} \right)^2} – {\left( {x + y} \right)^2}\\
= \left( {xy + 1 – x – y} \right)\left( {xy + 1 + x + y} \right)\\
= \left[ {\left( {xy – x} \right) – \left( {y – 1} \right)} \right].\left[ {\left( {xy + x} \right) + \left( {y + 1} \right)} \right]\\
= \left[ {x\left( {y – 1} \right) – \left( {y – 1} \right)} \right].\left[ {x\left( {y + 1} \right) + \left( {y + 1} \right)} \right]\\
= \left( {x – 1} \right)\left( {y – 1} \right)\left( {x + 1} \right)\left( {y + 1} \right)\\
e,\\
{x^2} – 7x + 12\\
= \left( {{x^2} – 3x} \right) – \left( {4x – 12} \right)\\
= x\left( {x – 3} \right) – 4\left( {x – 3} \right)\\
= \left( {x – 3} \right)\left( {x – 4} \right)\\
f,\\
{x^2} – 5x – 14\\
= \left( {{x^2} – 7x} \right) + \left( {2x – 14} \right)\\
= x\left( {x – 7} \right) + 2\left( {x – 7} \right)\\
= \left( {x – 7} \right)\left( {x + 2} \right)\\
g,\\
4{x^2} – 3x – 1\\
= \left( {4{x^2} – 4x} \right) + \left( {x – 1} \right)\\
= 4x\left( {x – 1} \right) + \left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {4x + 1} \right)
\end{array}\)