Sửa đề: \(x^4-x^2+2x-1\)

Ta có: \(x^4-x^2+2x-1\)

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

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

Đặt x²+x+1=t

Ta có: t(t+1)-12 = t²+t-12 = t²+4t-3t-12 = t(t+4) -3(t+4) =(t-3)(t+4) = (x²+x+1-3)(x²+x+1+4) =(x²+x-2)(x²+x+5).

= (x²+x+5)(x-1)(x+2)

x⁷+x²+1

=x²(x⁵+1)+1

x⁸+x+1

=x(x⁷+1)+1

Em sửa lại tên đi nhé!

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

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

= \(\left(x^2-1-\frac{x}{2}\right)^2-\frac{9}{4}x^2\)

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

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

Phân tích tiếp được đấy:

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

\(x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=\left(x-\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\)

Thay vào nhé!

(x²+x+1)(x²+x+2)-12

=\(\left(x^2+x+\frac{3}{2}-\frac{1}{2}\right)\left(x^2+x+\frac{3}{2}+\frac{1}{2}\right)-12\)
=\(\left(x^2+x+\frac{3}{2}\right)^2-\frac{1}{4}-12\)
=\(\left(x^2+x+\frac{3}{2}\right)^2-\frac{49}{4}\)
=\(\left(x^2+x+\frac{3}{2}-\frac{7}{2}\right)\left(x^2+x+\frac{3}{2}+\frac{7}{2}\right)\)

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

\(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt:   \(x^2+x+1=t\)    Khi đó ta có:

\(A=t\left(t+1\right)-12\)

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

Thay trở lại đc:

\(A=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)

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

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

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

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

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

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

\(x^5-x^3+x^2-1\)

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

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

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

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

Đặt x^2 + x+  1 = a => x^2 + x + 2 =a + 1 

Thay vòa ta có :

a( a+ 1 ) - 12 = a^2  + a - 12 = a^2 + 4a - 3a - 12 

=a (a+4) - 3 ( a+ 4 )

= ( a- 3 )(a+4)

Thây x^2 + x + 1 = a vào ta có 

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

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

đặt t=x²+x+1 ta được:

t.(t+1)-12

=t²-t-12

=t²+3x-4t-12

=t.(t+3)-4.(t+3)

=(t+3)(t-4)

thay t=x²+x+1 ta được:

(x²+x+4)(x²+x-3)

vậy (x² + x + 1) . (x² + x + 2) - 12=(x²+x+4)(x²+x-3)