Đáp án:
a, Ta có :
$(x - 2)(x^2 + 2x + 4) - x(x^2 - 3) = 7$
$<=> x^3 - 8 - x^3 + 3x = 7$
$ <=> 3x - 8 = 7$
$ <=> 3x = 15$
$ <=> x = 5$
b, Ta có :
$x(x + 1)^1 - ( x - 1)^3 - 5x(x + 2) = 9$
$ <=> x.(x^2 + 2x + 1) - (x^3 - 3x^2 + 3x - 1) - (5x^2 + 10x) = 9$
$ <=> x^3 + 2x^2 + x - x^3 + 3x^2 - 3x + 1 - 5x^2 - 10x = 9$
$ <=> ( x^3 - x^3) + (2x^2 + 3x^2 - 5x^2) - ( 10x + 3x - x) + 1 = 9$
$ <=> 12x = 8$
$ <=> x = 2/3$
c, Ta có :
$( x + 3)^3 - x(3x + 1)^2 + (2x + 1)(4x^2 - 2x + 1) = 28$
$ <=> x^3 + 9x^2 + 27x + 27 - x(9x^2 + 6x + 1) + (2x)^3 + 1^3 = 28$
$ <=> x^3 + 9x^2 + 27x + 27 - 9x^3 - 6x^2 - x + 8x^3 + 1 = 28$
$ <=> ( x^3 - 9x^3 + 8x^3) + (9x^2 - 6x^2) + (27x - x) + 28 = 28$
$ <=> 3x^2 + 26x = 0$
$ <=> x . (3x + 26) = 0$
<=> \(\left[ \begin{array}{l}x=0\\x=3x + 26 = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=0\\x=-26/3\end{array} \right.\)
d, Ta có :
$( x - 3)^3 - (x - 3)(x^2 + 3x + 9) + 6(x + 1)^2 + 3x^2 = -33$
$<=> x^3 - 9x^2 + 27x - 27 - (x^3 - 27) + 6(x^2 + 2x + 1) + 3x^2 = -33$
$ <=> x^3 - 9x^2 + 27x - 27 - x^3 + 27 + 6x^2 + 12x + 6 + 3x^2 = -33$
$ <=> 39x + 6 = -33$
$ <=> 39x = -39$
$ <=> x = -1$
Giải thích các bước giải:
