x⁴ - 12x² + 12x - 9 = (x⁴ - 3x³) + (3x³ - 9x²) + (- 3x² + 9x) + (3x - 9)

= (x - 3)(x³ + 3x² - 3x + 3) = 0

a) x⁴-12x³+12x-9=(x⁴-3x³)+(3x³-9x²)-(3x²-9x)+(3x-9)=x³(x-3)+3x²(x-3)-3x(x-3)+3(x-3)

=(x-3)(x³+3x²-3x+3)

b)P(x)=o=>x-3=0 và x³=3x²-3x+3=0

                 =>x=3 và x=rỗng

=>x=3

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

Đặt \(t=x^2-11x+30\)

\(\Rightarrow\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=t.\left(t-2x\right)-24x^2\)

\(=t^2-2xt-24x^2\)

\(=\left(t^2-2xt+x^2\right)-25x^2\)

\(=\left(t-x\right)-\left(5x\right)^2\)

\(=\left(t-6x\right)\left(t+4x\right)\)

\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)

Tham khảo nhé~

x^4-5x^2+4=x^4-x^2-(4x^2-4) = x^2(x^2-1)-4(x^2-1)

=(x^2-4)(x^2-1)

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

\(x^4-5x^2+4\)

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

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

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

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

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

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

a, x^4+6x^3+11x^2+6x+1 
= x^4 + 6x^3 + 9x² + 2x² + 6x + 1 
= x^4 + 9x² + 1 + 6x^3 + 2x² + 6x 
= x^4 + 9x² + 1² + 2.x².3x + 2.x².1 + 2.3x.1
= (x² + 3x + 1)²

Mình làm được ý a nên tk 1 tk

a, x²-7x-14y+2x

=x(x+2)-7(x-2y)

b, x³-4x²y+4xy²-25x

=x³-4x²y+4xy²-y³-25x+y³

=(x-y)³-25x+y³

a ) = x(x+2) - 7(x+2y)

b) =  -4 xy ( x-y) + (x^3-25x)  [ câu này mk , chaqcs là làm đúng đâu ]

a) ( x-2 )( x - 4 )(  x - 6 )( x  -8  ) + 15 

= ( x-  2 )( x - 8 )( x -  4)( x- 6  ) + 15

= ( x^2 - 10x + 16 )( x^2 - 10x + 24 ) + 15 

Đắt x^2 + x + 16 = y 

= y ( y + 8 ) + 15

= y^2 + 8y + 15

=  y^2 + 3y + 5y  + 15

=y ( y + 3 ) + 5 ( y + 3 )

= ( y+  5)( y + 3)

Thay vào