\(=x^8-3x^4+3x^2-1-x^8+1=-3x^4+3x^2\)

(x² - 1)³ - (x² - 1)(x⁴ + x² + 1)

= (x⁴ - 2x² + 1 - x⁴ - x² - x)(x² - 1)

= (-3x² - x + 1)(x² - 1)

= -3x⁴ + 3x² - x³ + x + x² - 1

= -3x⁴ + 4x² - x³ + x - 1

x(2 x 2  – 3) –  x 2 (5x + 1) +  x 2

= x. 2 x 2  + x.(- 3) – ( x 2 . 5x +  x 2  .1) +  x 2

= (2 x 3  – 3x) – (5 x 3  +  x 2 ) +  x 2

= 2 x 3  – 3x – 5 x 3  –  x 2  +  x 2

= -3x – 3 x 3

\(a,\dfrac{x+2}{x^2-4}=\dfrac{\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\\ b,\dfrac{x^2-9}{3-x}=\dfrac{\left(x-3\right)\left(x+3\right)}{-\left(x-3\right)}=-x-3\)

\(\left(x^2+2xy\right)^2+2\left(x^2+2xy\right)y^2+y^4\)

\(=\left(x^2+2xy+y^2\right)^2\)

\(=\left(x+y\right)^4\)

a) \(ĐKXĐ:x\ne\pm1\)

b) \(M=\dfrac{x^2+2x+1}{x^2-1}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)

b) hình như bạn viết nhầm đề 

x² + 4x + 3 = 0 

<=> (x + 1)(x + 3) =0

<=> \(\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)

Thay x = -3 vào P ta được :

\(P=\dfrac{-3+1}{-3-1}=\dfrac{-2}{-4}=\dfrac{1}{2}\)

vậy ...

\(A=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x=x^2+4x\\ B=\left(x^2+xy+y^2\right)\left(x-y\right)=x^3-y^3\)

a) 9x²+6xy+y²

= (3x)²+2.3x.y+y²

=(3x+y)²

b) x² + 4xy + y² 

=x² + 2xy + y² + 2xy

= (x+y)²+2xy   

\(a.9x^2+6xy+y^2\\ =\left(3x+y\right)^2\\ b.x^2+4xy+y^2\\ =\left(x+y\right)^2+2xy\)

c) x²+6xy+9y²

= (x+3y)²