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

\(=\left(x^2+x-1\right)^2+4\left(x^2+x-1\right)+3\)

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

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

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

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

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

Chúc bạn học tốt.

    x²+x-30

=x²-5x+6x-30

=x(x-5)+6(x-5)

=(x+6)(x-5)

a)

x²-4x+4-x²+5x=13

x+4=13

x=9

a)²(x-3)+12-4x

=x²(x-3)-4(x-3)

=(x²-4)(x-3)

=(x²-2²)(x-3)

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

b)x³-4x²-12x+27

=x³-7x²+9x+3x²-21x+27

=x(x²-7x+9)+3(x²-7x+9)

=(x+3)(x²-7x+9)

a)\(x^2\left(x-3\right)+12-4x\)

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

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

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

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

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

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

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

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\\ =\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

=\(x^4+2x^3+x^2+4x^2+4x-12\)

=\(x^4+2x^3+5x^2+4x-12\)

=\(x^4-x^3+3x^3-3x^2+8x^2+4x-12\)

=\(x^3(x-1)+3x^2(x-1)+4(2x^2+x-3)\)

=\(x^3(x-1)+3x^2(x-1)+4(2x^2-2x+3x-3)\)

=\(x^3(x-1)+3x^2(x-1)+4[2x(x-1)+3(x-1)]\)

=\(x^3(x-1)+3x^2(x-1)+4(x-1)(2x+3)\)

=\((x-1)[x^3+3x^2+4(2x+3)]\)

=\((x-1)(x^3+3x^2+8x+12)\)

\(\left(x^2+x\right)^2+\left(4x^2+4x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)

đặt y=x²+4x+8 ta được

y²+3xy+2x²=y²+xy+2xy+2x²=y(y+x)+2x(y+x)

=(y+x)(y+2x)

thay y=x²+4x+8 ta được

(x²+5x+8)(x²+7x+8)

=(x^2+4x+8)²+2x(x^2+4x+8)+(x^2+4x+8)+2x^2

=(x^2+5x+8)(x^2+6x+8)

(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2

dat x^2+4x+8=y

ta co:y^2+3xy+2x^2

=y^2+xy+2xy+2x^2

=y(y+x)+2x(y+x)

=(y+2x)(y+x)

=(x^2+4x+8+2x)(x^2+4x+8+x)

=(x^2+6x+8)(x^2+5x+8)

KL:......................

Lời giải:

a. 

$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

a, x^2-4x+3

=x^2-x-3x+3

=x(x-1)-3(x-1)

=(x-3)(x-1)

\(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)

\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)

\(=\left(x-1\right).\left(x^3+3x^2+8x+12\right)=\left(x-1\right).\left(x+2\right).\left(x^2+x+6\right)\)

p/s: sai sót bỏ qua