Đáp án:
$1)
c) 3a^4+3a^2\\
d) -54-27a^2-9a^4\\
2)
b) x=1\\
3)
-30$
Giải thích các bước giải:
$1)
c) (a^2 – 1)^3 – ( a^4 + a^2 + 1)(a^2 – 1)\\
=(a^2)^3-3a^4+3a^2-1-\left [(a^2)^3-1^3 \right ]\\
=a^6-3a^4+3a^2-1-a^6+1\\
=3a^4+3a^2\\
d) ( a^4 – 3a^2 + 9)( a^2 + 3) – ( 3 + a^2)^3\\
=(a^2)^3-3^3-(27+27a^2+9a^4+a^6)\\
=a^6-27-27-27a^2-9a^4-a^6\\
=-54-27a^2-9a^4\\
2)
b) (x – 2)^3 – (x – 3)(x^2 + 3x + 9) + 6(x + 1)^2 = 49\\
\Leftrightarrow x^3-6x^2+12x-8-(x^3-3^3)+6(x^2+2x+1)-49=0\\
\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6-49=0\\
\Leftrightarrow 24x-24=0\\
\Leftrightarrow 24x=24\\
\Leftrightarrow x=1\\
3)
(x – 1)^3 – 4x(x + 1)(x – 1) + 3(x – 1)(x^2 + x + 1)\\
=x^3-3x^2+3x-1-4x(x^2-1)+3(x^3-1)\\
=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\\
=-3x^2+7x-4$
Thay $x=-2$ ta được
$-3.(-2)^2+7.(-2)-4=(-3).4-14-4=-30$
4)
$(x – 1)^3 – (x – 1)(x^2 + x + 1) + 3(x – 1)x\\
=x^3-3x^2+3x-1-(x^3-1)+3x^2-3x\\
=x^3-3x^2+3x-1-x^3+1+3x^2-3x\\
=(x^3-x^3)+(-3x^2+3x^2)+(3x-3x)+(-1+1)
=0$