x^3+3x^2-7x^2-21x+9x+27=0

x^2(x+3)-7x(x+3)+9(x+3)=0

(x+3)(x^2-7x+9)=0

x = -3 hoặc x^2 - 7x + 9 = 0 (chuyển về pt dạng kx^2 + m)

Bạn chuyển 7x = 2 . x . 7/2 + 49/4 - 49/4

=x^3 +3x^2 - 7x^2 - 21x +9x + 27 

=x^2(x+3)-7x(x+3)+9(x+3) 

=(x+3)(x^2-7x+9)

(𝑥+3)(𝑥2−7𝑥+9)

Với x = -3 ta có -27-4*9+ 36+27=0 do đó đa thức chứa nhân tử x+3
Ta có: x^3 -4x^2-12x+27 = x^3 +3x^2 -7x^2-21x+9x+27 =(x^3 +3x^2)-(7x^2+21x) + (9x+27) =x^2(x+3) -7x(x+3)+ 9(x+3)=(x+3)(X^2 - 7x+9)
* Xét x^2 -7x + 9 = x^2 - 2x.7/2 +49/4-49/4+9 = (x-7/2)^2 -13/4 =(x-7/2- √13/2)(x-7/2+√13/2)
Vậy: x^3 -4x^2-12x+27 = (x+3)(x-7/2)^2 -13/4 =(x-7/2- √13/2)(x-7/2+√13/2)

c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)

\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)

d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)

Tương tự với a ; b 

a)\(4x^2-4x+4=0\Leftrightarrow\left(2x-1\right)^2+3\) (đến đây hết pt dc rùi)

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

c)\(x^3-4x^2+3x=x^3-x^2-3x^2+3x\)

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

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

d)\(4x^2-12x+3=\left(2x-3\right)^2-6\)

                                    =\(\left(2x-3\right)^2-\sqrt{6^2}\)

                                 =\(\left(2x-3-\sqrt{6}\right)\left(2x-3+\sqrt{6}\right)\)

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

\(b,x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)

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

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

                             \(=x\left[x\left(x-1\right)-3\left(x-1\right)\right]\)

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

\(d,4x^2-12x+3=4\left(x^2-3x+\frac{3}{4}\right)\)

\(=4\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}+\frac{3}{4}\right)\)

\(=4\left[\left(x-\frac{3}{2}\right)^2-\frac{3}{2}\right]\)

\(=4\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2\right]\)

\(=4\left(x-\frac{3}{2}-\frac{\sqrt{3}}{\sqrt{2}}\right)\left(x-\frac{3}{2}+\frac{\sqrt{3}}{\sqrt{2}}\right)\)

\(=4\left(x-\frac{3+\sqrt{6}}{2}\right)\left(x-\frac{3-\sqrt{6}}{2}\right)\)

P/s: Dương: câu d t k chắc nx, sai thì thông cảm :)) -Huyền Nhi-

Giúp mình vs mình đang gấp

\(=x^3+3x^2-7x^2-21x+9x+27=\left(x+3\right)\left(x^2-7x+9\right)\)

Đáp án:
   x³-4x²-12x+27
= (x+3)(x²-7x+9)

a) `x^4+2x^3-4x-4`

`=(x^4-4)+(2x^3-4x)`

`=(x^2-2)(x^2+2)+2x(x^2-2)`

`=(x^2-2)(x^2+2+2x)`

b) `x^3-4x^2+12x-27`

`=(x^3-27)-(4x^2-12x)`

`=(x-3)(x^2+3x+9)-4x(x-3)`

`=(x-3)(x^2+3x+9-4x)`

`=(x-3)(x^2-x+9)`

c) `xy-4y-5x+20`

`=y(x-4)-5(x-4)`

`=(y-5)(x-4)`

a) Ta có: \(x^4+2x^3-4x-4\)

\(=\left(x^4-4\right)+2x^3-4x\)

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

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

b) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\cdot\left(x-3\right)\)

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

c) Ta có: \(xy-4y-5x+20\)

\(=y\left(x-4\right)-5\left(x-4\right)\)

\(=\left(x-4\right)\left(y-5\right)\)

câu trả lời là:

12x+27=39-4=35x3x2=6+0=6

đáp án là 6