1)tìm x
a) (x-2)^2-x(x+3)-4x=-17
b)(x-2)^2-(x+5)(x-5)=0
c)5x^3-10x^2+5x=0
2)phân tích thành nt
8x^3+12x^2+6x+1-(x+1)^3
1)tìm x
a) (x-2)^2-x(x+3)-4x=-17
b)(x-2)^2-(x+5)(x-5)=0
c)5x^3-10x^2+5x=0
2)phân tích thành nt
8x^3+12x^2+6x+1-(x+1)^3
Đáp án:
1, Tìm x
a) $(x-2)^{2}$ – x(x+3) – 4x = -17
<=> $x^{2}$ – 4x + 4 – $x^{2}$ – 3x – 4x = -17
<=> -11x = -21
<=> x = $\frac{21}{11}$
b)$(x-2)^{2}$ – (x+5)(x-5) = 0
<=> $x^{2}$ – 4x + 4 – $x^{2}$ + 25 = 0
<=> -4x = -25
<=> x = $\frac{25}{4}$
c) $5x^{3}$ – $10x^{2}$ + 5x = 0
<=> x($5x^{2}$ – 10x + 5) = 0
<=> x($5x^{2}$ – 5x – 5x + 5) = 0
<=> 5x($(x-1)^{2}$) = 0
<=> \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
2, PTĐTTNT
$8x^{3}$ + $12x^{2}$ + 6x + 1 – $(x-1)^{3}$
= $8x^{3}$ + $12x^{2}$ + 6x + 1 – $x^{3}$ – $3x^{2}$ – 3x – 1
= $7x^{3}$ + $9x^{2}$ + 3x = x($7x^{2}$ + 9x + 3)
Giải thích các bước giải:
$1$)
$a$) `(x-2)^2 – x(x+3) – 4x = -17`
`⇔ x^2 – 4x +4 – x^2 – 3x – 4x = -17`
`⇔ -11x + 4 = -17`
`⇔ -11x = -21`
`⇔ x = 21/11`
Vậy `x=21/11`.
$b$) `(x-2)^2-(x+5)(x-5)=0`
`⇔ x^2 – 4x + 4 – x^2 + 25 = 0`
`⇔ -4x + 29 =0 `
`⇔ x = 29/4`
Vậy `x=29/4`.
$c$) $5x^3 – 10x^2 + 5x = 0$
$⇔ 5x^3 – 5x^2 – 5x^2 + 5x = 0$
$⇔ 5x^2(x-1) – 5x(x-1) = 0$
$⇔ 5x(x-1)^2 = 0$
$⇒$ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy $x$ $∈$ `{0;1}`.
$2$)
`8x^3+12x^2+6x+1-(x+1)^3`
`= 8x^3 + 12x^2 + 6x + 1 – (x^3 + 3x^2 + 3x + 1)`
`= 8x^3 + 12x^2 + 6x + 1 – x^3 – 3x^2 – 3x – 1`
`= 7x^3 + 9x^2 + 3x`
`= x(7x^2 + 9x+3)`.