c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2=0\)

hay x=1

b) 4x(2-x)+(2x+1)^2=2
    8x-4x^2+4x^2+4x+1-2=0
    (8x+4x)+(-4x^2+4x^2)+(1-2)=0
         12x + 0 -1 =0
                 12x=1
                     x=1/12
Vậy x= 1/2
c) (x-3)^3-x^2(x-9)=0
     x^3-9x^2+27x-x^3+9x^2=0
     (x^3-x^3)+(-9x^2+9x^2)+27x=0
           0 + 0 + 27x=0
                     x= 0
Vậy x=0
    

\(a,=12-3x+4x-x^2+x^2-2x=12-x\\ b,=x^2-2x+1-x^2+4=-2x+5\)

\(a,=x^2+4x+4\\ b,=x^3+3x^2+3x+1\\ c,=\left(x-3\right)\left(x+3\right)\)

a,\(\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+4\)

b, \(\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\)

c,\(x^2-3^2=\left(x-3\right).\left(x+3\right)\)

a,

b, 

c,

\(\Rightarrow8x-4x^2+4x^2+4x+1=2\\ \Rightarrow12x=1\Rightarrow x=\dfrac{1}{12}\)

4x(2-x)+(2x+1)²=2

4x2-4xx+(2x)²+2.2x1+1²=2

8x-4x²+4x²+4x+1=2

12x+1=2

12x=2-1

12x=1

x=1/12

từ từ sửa lại

x(x + 3) – x^2 = 6. 

oki

^2 nha

\(\Leftrightarrow8x-4x^2+4x^2+4x+1=2\)

\(\Leftrightarrow x=\dfrac{1}{12}\)

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

\(\Leftrightarrow8x-4x^2+4x^2+4x+1=2\)

\(\Leftrightarrow12x=1\)

\(\Leftrightarrow x=\dfrac{1}{12}\)

b: \(=x^2+6x+9-x^2+4=6x+13\)

\(a,\left(x-2\right)^2=x^2-2.x.2+2^2=x^2-4x+4\\ b,\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\\ c,x^2-5x^2=-4x^2\)