\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)

Xem lại đề ý b

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

b. \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

a: \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)

b: \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

x²-6xy+9y²-36 =(x²-6xy+9y²)-36 =(x-3y)²-6² =(x-3y+6)(x-3y-6)

\(3x^2-6xy+9y^2-12\)

\(=3\cdot x^2-3\cdot2xy+3\cdot3y^2-3\cdot4\)

\(=3\cdot\left(x^2-2xy+3y^2-4\right)\)

=3*x^2-3*2xy+3*3y^2-3*4

=3(x^2-2xy+3y^2-4)

a) \(2x^3+6xy-x^2z-3yz\)

= \(\left(2x^3+6xy\right)-\left(x^2z+3yz\right)\)

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

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

b)\(x^2-6xy+9y^2-49\)

=\(x^2-2.x.3y+\left(3y\right)^2-7^2\)

=\(\left(x-3y\right)^2-7^2\)

=\(\left(x-3y+7\right)\left(x-3y-7\right)\)

Cứu với mn ơi , mai em có ktra ạ 

x^2+6xy+9y^2-3x-9y+2

=( x^2+6xy+9y^2)-3(x+3y)+9/4 -1/4

=(x+3y)^2-3(x+3y)+(3/2)^2- 1/4

=(x+3y+3/2)^2-(1/2)^2

=(x+3y+3/2+1/2)(x+3y+3/2-1/2)=(x+3y+2)(x+3y+1)

\(x^4+6x^3+13x^2+12x+4\)

\(=x^4+x^3+5x^3+5x^2+8x^2+8x+4x+4\)

\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+8x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+5x^2+8x+4\right)\)

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

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

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

=[x²-6xy+(3y)²-49]

=[(x-3y)²-7²]

=(x-3y-7)*(x+3y+7)

= x^2 - 6xy + 9y^2 - 49

= x^2 - 6xy + ( 3y )^2 - 49

= (x - 3y)^2 - 7^2

= (x - 3y -7)( x-3y +7) tk cho mk nha

a) = (x - 2)² - y²

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

b) = (x^2 + 6x + 9) - (2y)^2

= (x + 3)² - (2y)²

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

c) = (x - 3y)² - 6²

= (x - 3y - 6)(x - 3y + 6)