Answer:

\(81x^4+4\)

\(=81x^4+36x^2+4-36x^2\)

\(=\left(9x^2+2\right)^2-6x^2\)

\(=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)

x²-6x+5

=x²-5x-x+5

=x(x-5)-(x-5)

=(x-5)(x-1)

\(x^2-6x+5\)

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

\(=x\left(x-1\right)-5\left(x-1\right)\)

\(=\left(x-1\right)\left(x-5\right)\)

\(x^2-x-12=x^2-x+\frac{1}{4}-\frac{1}{4}-12=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}=\left(x+3\right)\left(x-4\right)\)

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

x^2-x-12

=x^2-x-16+4

=x^2-16-x+4

=(x-4)(x+4)-(x-4)

=(x-4)(x+3)

(1+x2)2−4x(1−x2)

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

đặt \(\left(1-x^2\right)\)= a

ta có :

- a . a - 4x .a

= a ( - a - 4x )

thay a = \(\left(1+x^2\right)\) ta có

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

phân tích tiếp nhé !
 

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

1/ Đặt x²+x=t

=>(x²+x)²+4(x²+x)-12=t²+4t-12=t²+6t-2t-12=t(t+6)-2(t+6)=(t-2)(t+6)=(x²+x-2)(x²+x+6)=(x²-x+2x-2)(x²+x+6)

=[x(x-1)+2(x-1)](x²+x+6)=(x-1)(x+2)(x²+x+6)

2/ Đặt x²+x=t 

=>(x²+x)²+9x²+9x+14=(x²+x)²+9(x²+x)+14=t²+9t+14=t²+2t+7t+14=t(t+2)+7(t+2)=(t+2)(t+7)=(x²+x+2)(x²+x+7)

3/ Đặt x²+5x=t

=>(x²+5x)²+10x²+50x+24=(x²+5x)²+10(x²+5x)+24=t²+10t+24=t²+4t+6t+24=t(t+4)+6(t+4)=(t+4)(t+6)=(x²+5x+4)(x²+5x+6)

=(x²+x+4x+4)(x²+2x+3x+6)=[x(x+1)+4(x+1][x(x+2)+3(x+2)]=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+2)(x+3)(x+4)

x(x-5)(x+2)

(x² - 8)² + 36 = x⁴ - 16x + 100 = x⁴ - 20x² + 100 - 4x² 

                      = (x² - 10) - 4x² = (x² - 10 - 2x)(x² - 10 + 2x)