phân tích đa thức thành nhân tử

e) x³-3x²-4x+12
= x²(x-3)-4(x-3)
= (x²-4)(x-3)

a) x³ - 64x = 0

x(x² - 64) = 0

x(x - 8)(x + 8) = 0

x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0

*) x - 8 = 0

x = 8

*) x + 8 = 0

x = -8

Vậy x = -8; x = 0; x = 8

b) x³ - 4x² = -4x

x³ - 4x² + 4x = 0

x(x² - 4x + 4) = 0

x(x - 2)² = 0

x = 0 hoặc (x - 2)² = 0

*) (x - 2)² = 0

x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) x² - 16 - (x - 4) = 0

(x - 4)(x + 4) - (x - 4) = 0

(x - 4)(x + 4 - 1) = 0

(x - 4)(x + 3) = 0

x - 4 = 0 hoặc x + 3 = 0

*) x - 4 = 0

x = 4

*) x + 3 = 0

x = -3

Vậy x = -3; x = 4

d) (2x + 1)² = (3 + x)²

(2x + 1)² - (3 + x)² = 0

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

(x - 2)(3x + 4) = 0

x - 2 = 0 hoặc 3x + 4 = 0

*) x - 2 = 0

x = 2

*) 3x + 4 = 0

3x = -4

x = -4/3

Vậy x = -4/3; x = 2

e) x³ - 6x² + 12x - 8 = 0

(x - 2)³ = 0

x - 2 = 0

x = 2

f) x³ - 7x - 6 = 0

x³ + 2x² - 2x² - 4x - 3x - 6 = 0

(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0

x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0

(x + 2)(x² - 2x - 3) = 0

(x + 2)(x² + x - 3x - 3) = 0

(x + 2)[(x² + x) - (3x + 3)] = 0

(x + 2)[x(x + 1) - 3(x + 1)] = 0

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

x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x + 1 = 0

x = -1

*) x - 3 = 0

x = 3

Vậy x = -1; x = -1; x = 3

a,x\(^3\)-64=0
   x\(^3\)     =64
  =>x=3
b,x\(^3\)-4x\(^2\)=-4x

   x\(^3\)-4x\(^2\)+4x=0
   x(x\(^2\)-4x+4)=0
   x(x-2)\(^2\)=)
TH1:x=0
TH2:x-2=0
     =>x=2

c,x\(^2\)-16-(x-4)=0

   (x+4)(x-4)-(x-4)=0
   (x-4)(x+4-1)=0

   (x-4)(x+3)=0
TH1:x-4=0
     =>x=4

TH2:x+3=0

    =>x=-3

d,(2x+1).2=3+x
    4x+2-3-x=0
    3x-1=0
     x=\(\dfrac{1}{3}\)

e,x\(^3\)-6x\(^2\)+12x-8=0
   (x-2)\(^3\)=0
=>x-2=0
=>x=2

f,x\(^3\)-7x+6=0
  x\(^3\)-x-6x+6=0
  x(x\(^2\)-1)-6(x-1)=0
  x(x+1)(x-1)-6(x-1)=0
  (x-1)(x\(^2\)+x-6)=0
TH1:x-1=0

  =>x=1
TH2:x\(^2\)+x-6=0

       x\(^2\)+3x-2x-6=0

       x(x+3)-2(x+3)=0

       (x+3)(x-2)=0

=>x+3=0                =>x-2=0

+>x=-3                   =>x=2

d,(2x+1)\(^2\)=(3+x)\(^2\)

4x\(^2\)+4x+1-9-6x-x\(^2\)=0
3x\(^2\)-2x-8=0
3x\(^2\)-6x+4x-8=0
3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

TH1:3x+4=0                     TH2:x-2=0
=>x=\(\dfrac{-4}{3}\)                              =>x=2

giúp mình nha

a)14𝑥^3𝑦∶10𝑥^2=1,4𝑥𝑦

b)(𝑥^3-27):(3-𝑥)=(𝑥-3)(𝑥^2+3𝑥+9):(3-𝑥)=-(𝑥^2+3𝑥+9)=-𝑥^2-3𝑥-9

c)8𝑥^3𝑦^3𝑧∶6𝑥𝑦^3=4/3𝑥^2𝑧

d)(𝑥^2−9𝑦^2+4𝑥+4)∶(𝑥+3𝑦+2)=((𝑥^2+4𝑥+2^2)-(3𝑦)^2):(𝑥+3𝑦+2)=((𝑥+2)^2-(3𝑦)^2):(𝑥+3𝑦+2)=(𝑥+2-3𝑦)(𝑥+2+3𝑦):(𝑥+3𝑦+2)=𝑥+2-3𝑦

a) \(14x^3y:10x^2=\dfrac{7}{5}xy\)

b) \(\left(x^3-27\right):\left(3-x\right)\)

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

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

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

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

bạn nào giải giúp mình nha

c) \(8x^3y^3z:6xy^3=\dfrac{4}{3}x^2z\)

d) \(\left(x^2-9y^2+4x+4\right):\left(x^2+3y+2\right)\)

\(=\left[\left(x+2\right)^2-\left(3y\right)^2\right]:\left(x^2+3y+2\right)\)

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

\(=x-3y+2\)

a) 

8x³y³z:6xy³=\(\dfrac{4}{3}\)x²z

d) (x²−9y²+4x+4):(x+3y+2)

=[(x+2⁾²−(3y)²]:(x+3y+2)

=(x+3y+2)(x−3y+2):(x+3y+2)

=x-3y+2

\(b,\Leftrightarrow x+7=38\Leftrightarrow x=31\\ c,\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\\ d,\Leftrightarrow2x=160-49=111\Leftrightarrow x=\dfrac{111}{2}\\ e,\Leftrightarrow x-8=20\Leftrightarrow x=28\\ f,\Leftrightarrow x-3=\dfrac{59}{4}\Leftrightarrow x=\dfrac{71}{4}\\ g,\Leftrightarrow x=3\\ h,\Leftrightarrow2x+1=5\Leftrightarrow2x=4\Leftrightarrow x=2\)