Question

Phân tích thành nhân tử :

     1.x³ + 2x² + 2x + 1

     2.x³ - 4x² + 12x - 27

     3. x⁴  - 2x³ + 2x - 1

     4.x⁴ + 2x³ + 2x² + 2x +1

Answer 1

1. \(x^3+2x^2+2x+1\)

\(=\left(x^3+1\right)+\left(2x^2+2x\right)\)

\(=\left(x+1\right)\left(x^2+x+1\right)+2x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+3x+1\right)\)

2. \(x^3-4x^2+12x-27\) 

\(=x^3-3x^2-x^2+3x+9x-27\)

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x^2-x+9\right)\left(x-3\right)\)

3. \(x^4-2x^3+2x-1\)

 \(=\left(x^4-1\right)-2x\left(x^2-1\right)\)

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

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

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

\(=\left(x-1\right)^3\left(x+1\right)\)

d) \(x^4+2x^3+2x^2+2x+1\)

\(=x^4+x^2+2x^3+2x+x^2+1\)

\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)

\(=\left(x^2+2x+1\right)\left(x^2+1\right)\)

\(=\left(x+1\right)^2\left(x^2+1\right)\)

Answer 2

1. x³ + 2x² + 2x + 1
= x³ + x² + x² + x + x + 1
= x²(x + 1) + x(x + 1) + (x + 1)
= (x + 1)(x² + x + 1)

2. x³ - 4x² + 12x - 27
= x³ - 3x² - x² + 3x + 9x - 27
= x²(x - 3) - x(x - 3) + 9(x - 3)
= (x - 3)(x² - x  + 9)

3. x⁴ - 2x³ + 2x - 1
= x⁴ - x³ - x³ + x² - x² + x + x - 1
= x³(x - 1) - x²(x - 1) - x(x - 1) + (x - 1)
= (x - 1)(x³ - x² - x + 1)
= (x - 1)[x(x² - 1) - (x² - 1)]
= (x - 1)(x² - 1)(x - 1)
= (x - 1)²(x - 1)(x + 1)
= (x - 1)³(x + 1)

4. x⁴ + 2x³ + 2x² + 2x + 1
= x⁴ + x³ + x³ + x² + x² + x + x + 1
= x³(x + 1) + x²(x + 1) + x(x + 1) + (x + 1)
= (x + 1)(x³ + x² + x + 1)
= (x + 1)[x(x² + 1) + (x² + 1)]
= (x + 1)(x + 1)(x² + 1)
= (x + 1)²(x² + 1)

Answer 3

x4+2x3−2x2+2x−3=0⇔x4+3x3−x3−3x2+x2+3x−x−3=0⇔x3(x+3)−x2(x+3)+x(x+3)−(x+3)=0⇔(x+3)(x3−x2+x−1)=0⇔(x+3)[x2(x−1)+(x−1)]=0⇔(x+3)(x−1)(x2+1)=0⇔⎡⎣⎢x+3=0x−1=0x2+1=0⇔[x=−3x=1(vì x2+1≥1>0)x4+2x3−2x2+2x−3=0⇔x4+3x3−x3−3x2+x2+3x−x−3=0⇔x3(x+3)−x2(x+3)+x(x+3)−(x+3)=0⇔(x+3)(x3−x2+x−1)=0⇔(x+3)[x2(x−1)+(x−1)]=0⇔(x+3)(x−1)(x2+1)=0⇔[x+3=0x−1=0x2+1=0⇔[x=−3x=1(vì x2+1≥1>0)

Vậy ...

(x−1)(x2+5x−2)−x3+1=0⇔(x−1)(x2+5x−2)−(x3−1)=0⇔(x−1)(x2+5x−2)−(x−1)(x2+x+1)=0⇔(x−1)[(x2+5x−2)−(x2+x+1)]=0⇔(x−1)(4x−3)=0⇔[x−1=04x−3=0⇔⎡⎣x=1x=34(x−1)(x2+5x−2)−x3+1=0⇔(x−1)(x2+5x−2)−(x3−1)=0⇔(x−1)(x2+5x−2)−(x−1)(x2+x+1)=0⇔(x−1)[(x2+5x−2)−(x2+x+1)]=0⇔(x−1)(4x−3)=0⇔[x−1=04x−3=0⇔[x=1x=34

Vậy ...

x2+(x+2)(11x−7)=4⇔x2−4+(x+2)(11x−7)=0⇔(x+2)(x−2)+(11x−7)=0⇔(x+2)(x−2+11x−7)=0⇔(x+2)(12x−9)=0⇔3(x+2)(4x−3)=0⇔[x+2=04x−3=0⇔⎡⎣x=−2x=34

Vậy.......