= 10² * x² - 1

= 100 * x² -1 

\(10x^2-1=\left(\sqrt{10}x-1\right)\left(\sqrt{10}x+1\right)\)

\(\left(10x\right)^2-1=\left(10x-1\right)\left(10x+1\right)\)

\(9a^2-6ab+b^2-1=\left(3a-b\right)^2-1^2=\left(3a-b-1\right)\left(3a-b+1\right)\)

9a^2 - 6ab + b^2 - 1= (3a)^2 - 2.3a.b + b^2 -1

                            = (3a - b)^2 - 1^2

                            = (3a - b - 1).(3a - b + 1)

  9a² - 6ab + b² - 1 = (3a - b)² - 1

                                   =  (3a - b + 1)(3a-b-1)

(4a⁴-12a²+1)=(4a⁴-8a²+1)-4a²=(2a²-1)²-4a²=(2a²-2a-1)(2a²+2a-1)

 4x^4 - 32x^2 +1 = 4x^4 + 4x^2 +1 - 36x^2 = (2x^2 + 1)^2 - 36x^2 = (2x^2 - 6x + 1)(2x^2 + 6x + 1) 

4 x⁴ - 32 x² + 1 

= ( 2 x² )² - 2 . 2x². 8 + 64 - 63 

= ( 2 x² - 8 )² - 63 

= ( 2x² - 8 + √63 ) ( 2x² - 8 - √63 )

Xong

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

=(x-y)(x+y) - (x+y)

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

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

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

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

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

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

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

x(x+2)(x^2+2x+2)+1 = (x^2+2x)(x^2+2x+1)+1

Đặt x^2+2x+1=y ta được:

(y-)(y+1)+1=y^2-1+1=y^2

= (x^2+2x+1)^2

= ( x + 1 )^4