a) \(9x^2-16\)

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

\(=\left(3x-4\right)\left(3x+4\right)\)

b) \(x^2+4xy+4y^2-3x-6y\)

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

\(=\left[x^2+2\cdot x\cdot2y+\left(2y\right)^2\right]-3\left(x+2y\right)\)

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

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

#\(Toru\)

a: \(=4xy\left(1-5x^2y\right)\)

b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)

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

d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)

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

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)

x² - 4xy + 4y² - z² + 2zt - t²

= (x² - 4xy + 4y²) - (z² - 2zt + t²)

= (x - 2y)² - (z - t)²

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

1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2, \(x^2-10x+25=\left(x-5\right)^2\) 

3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2) \(x^2-10x+25=\left(x-5\right)^2\)

3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)

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

1: \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2: \(x^2-10x+25=\left(x-5\right)^2\)

3: \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)

56.B

57.B

58.B

56B

57B

58B

Câu 1:

$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$

$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$

Câu 2:

$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$

$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$

Câu 1:

\(x^2+4y^2+4xy-16\)

\(=\left(x+2y\right)^2-16\)

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

Câu 2:

\(x^3+x^2+y^3+xy\)

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

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

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

C1:x^2+4y^2+4xy-16

=[x^2+4xy+(2y)^2]-16

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

=(x+2y-4)(x+2y+4)

C2: x^3+x^2+y^3+xy

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

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

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

bài này ra lâu r nhưng ngứa tay nên giải luôn=)))))