x⁴+4

= x⁴+4x²+4-4x²

=(x⁴+4x²+4)-4x²

=[ (x²)²+2.2x²+2²]-(4x)²

= (x²+2)²-(2x)²

=(x²+2-2x)(x²+2+2x)

chúc bạn học tốt

x²+7x+12

=x²+4x+3x+12

=x(x+4)+3(x+4)

= (x+3)(x+4)

e (3x+2)\(⋮\) (x+1)

vì (x+1)\(⋮\) (x+1)

=> (3x+3)\(⋮\) (x+1)

=> (3x+2)-(3x+3)\(⋮\) (x+1)

=>(3x+2-3x-3)\(⋮\) (x+1)

=> -1\(⋮\) (x+1)

=> (x+1)\(\in\) Ư(-1)={-1;1}

ta có bảng sau

vậy x=0

c x³-4x²+12x-27

= x³-3x²-x²+3x+9x-27

= (x³-3x²)-(x²-3x)+(9x-27)

=x²(x-3)-x(x-3)+9(x-3)

=(x-3)(x²-x+9)

b x³+2x²+2x+1

= x³+x²+x²+x+x+1

= (x³+x²⁾+(x²+x)+(x+1)

=x²(x+1)+x(x+1)+(x+1)

= (x+1)(x²+x+1)

x²-3xy-10y²

=x²+2xy-5xy-10y²

=x(x+2y)-5y(x+2y)

=(x-5y)(x+2y)

(x²+y²+2xy-1)(x²+y²+2xy-4)+2

đặt x²+y²+2xy-1=a ta có

a(a-3)+2

=a²-3a+2

=a²-a-2a+2

=(a²-a)-(2a-2)

=a(a-1)-2(a-1)

=(a-1)(a-2)

thay a=x²+y²+2xy-1 vào ta có

(x²+y²+2xy-1-1)(x²+y²+2xy-1-2)

=(x²+y²+2xy-2)(x²+y²+2xy-3)

bài 2

a) ( x+2)²-9=0

=>(x+2)²- 3² =0

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

=>(x-1)(x+5)=0

=>\(\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)