Question

[eim hân kawaii~]

tìm x:+)x^3-3x^2+2x+1=0

+)-x^3+7x^2-3x-3=0

+)x^3-7x+6x=0

+)x^4+3x^3-8x+3x+1=0

hộ đứa trẻ lop1 với ạ...

Answer 1

[Đặng Thị Thảo Vân]

2)

$$-x^3 + 7x^2 - 3x - 3 = 0$$$$\Leftrightarrow x^3 - 7x^2 + 3x + 3 = 0$$$$\Leftrightarrow (x^3 - x^2) - (6x^2 - 6x) - (3x - 3) = 0$$$$\Leftrightarrow x^2(x - 1) - 6x(x - 1) - 3(x - 1) = 0$$$$\Leftrightarrow (x - 1)(x^2 - 6x - 3) = 0$$$$\Leftrightarrow \left[\begin{array}{l} x - 1 = 0 \\ x^2 - 6x - 3 = 0 \end{array}\right.$$$$\Leftrightarrow \left[\begin{array}{l} x = 1 \\ (x - 3)^2 = 12 \end{array}\right.$$$$\Leftrightarrow \left[\begin{array}{l} x = 1 \\ x - 3 = \pm 2\sqrt{3} \end{array}\right.$$$$\Leftrightarrow \left[\begin{array}{l} x = 1 \\ x = 3 \pm 2\sqrt{3} \end{array}\right.$$
3)$$x^3 - 7x + 6 = 0$$$$\Leftrightarrow x^3 - x - 6x + 6 = 0$$$$\Leftrightarrow x(x^2 - 1) - 6(x - 1) = 0$$$$\Leftrightarrow x(x - 1)(x + 1) - 6(x - 1) = 0$$$$\Leftrightarrow (x - 1)\left[x(x + 1) - 6\right] = 0$$$$\Leftrightarrow (x - 1)(x^2 + x - 6) = 0$$$$\Leftrightarrow (x - 1)(x^2 - 2x + 3x - 6) = 0$$$$\Leftrightarrow (x - 1)\left[x(x - 2) + 3(x - 2)\right] = 0$$$$\Leftrightarrow (x - 1)(x - 2)(x + 3) = 0$$$$\Leftrightarrow \left[\begin{array}{l} x - 1 = 0 \\ x - 2 = 0 \\ x + 3 = 0 \end{array}\right. \Leftrightarrow \left[\begin{array}{l} x = 1 \\ x = 2 \\ x = -3 \end{array}\right.$$
4)$$x^4 + 3x^3 - 8x^2 + 3x + 1 = 0$$$$\Leftrightarrow x^2\left(x^2 + 3x - 8 + \frac{3}{x} + \frac{1}{x^2}\right) = 0$$$$\Leftrightarrow \left(x^2 + \frac{1}{x^2}\right) + 3\left(x + \frac{1}{x}\right) - 8 = 0 \quad (\text{với } x \neq 0)$$$$\text{Đặt } t = x + \frac{1}{x} \implies t^2 = x^2 + 2 + \frac{1}{x^2} \implies x^2 + \frac{1}{x^2} = t^2 - 2$$$$(t^2 - 2) + 3t - 8 = 0$$$$\Leftrightarrow t^2 + 3t - 10 = 0$$$$\Leftrightarrow t^2 - 2t + 5t - 10 = 0$$$$\Leftrightarrow t(t - 2) + 5(t - 2) = 0$$$$\Leftrightarrow (t - 2)(t + 5) = 0$$$$\Leftrightarrow \left[\begin{array}{l} t = 2 \\ t = -5 \end{array}\right.$$

• Với $t = 2$:
  $$x + \frac{1}{x} = 2 \Leftrightarrow \frac{x^2 - 2x + 1}{x} = 0 \Leftrightarrow (x - 1)^2 = 0 \Leftrightarrow x = 1$$
• Với $t = -5$:$$x + \frac{1}{x} = -5 \Leftrightarrow \frac{x^2 + 5x + 1}{x} = 0 \Leftrightarrow x^2 + 5x + 1 = 0$$$$\Delta = 5^2 - 4(1)(1) = 21 \implies x = \frac{-5 \pm \sqrt{21}}{2}$$

Answer 2

[Nguyễn Lê Phước Thịnh]

b: \(-x^3+7x^2-3x-3=0\)

=>\(x^3-7x^2+3x+3=0\)

=>\(x^3-x^2-6x^2+6x-3x+3=0\)

=>\(\left(x-1\right)\left(x^2-6x-3\right)=0\)

TH1: x-1=0

=>x=1

TH2: \(x^2-6x-3=0\)

=>\(x^2-6x+9-12=0\)

=>\(\left(x-3\right)^2=12\)

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3+2\sqrt3\\ x=3-2\sqrt3\end{array}\right.\)

c: \(x^3-7x+6=0\)

=>\(x^3-x-6x+6=0\)

=>x(x-1)(x+1)-6(x-1)=0

=>(x-1)[x(x+1)-6]=0

=>\(\left(x-1\right)\left(x^2+x-6\right)=0\)

=>(x-1)(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\\ x=2\end{array}\right.\)

Answer 3

[Trần Thị Thùy Dương]

lp 1 hc bài này???

Answer 4

[Trần Thị Thùy Dương]

Please, don't let it get classified as "Class 1" again—okay? I don't want to keep repeating myself; consider this a final warning: don't let it happen again. If it gets marked as "Class 1" but you believe that's incorrect, you must ensure it is corrected to the right level!

Answer 5

[eim hân kawaii~]

kh thấy toán lớp 8 ạ?mấy bé mặc định ngkhc lop1^^

Answer 6

[Trần Thị Thùy Dương]

ko thấy đó, thì sao, mấy bé để mik lp 1

Answer 7

[eim hân kawaii~]

um mù...

Answer 8

[eim hân kawaii~]

để lop1 thì s chet ai?dở hơi 215 giờ có mấy bn ngáo v à?

Answer 9

[Trần Thị Thùy Dương]

dở hơi, h có mấy đứa thick để lp 1 ah, ngta lại nói cho rồi nói ko ph

Answer 10

[eim hân kawaii~]

thì kh ph tht thì nói kh ph tht kh lẽ nhận mih lop1 ngu à?

Answer 11

[eim hân kawaii~]

tốt nhất im kh chet ai đâu dở d chịu đc

Answer 12

[Trần Thị Thùy Dương]

thì m cứ thck thì m cứ đi đi, t ko cấm

Answer 13

[Chu Thị Mai Hương]

Mấy con tuat chửi nhau à