(x+5)²

\(x^2+10x+25=\left(x+5\right)^2\)

\(\dfrac{1}{4}x^2+2xy+4y^2=\left(\dfrac{1}{2}x+2y\right)^2\)

\(x^2-25-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-25\)

\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)

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

\(=\left(x-2y-5\right)\cdot\left(x-2y+5\right)\)

a) \(4x^2-4xy+y^2-9\)

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

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

b) \(x^2-36+4xy+4y^2\)

\(=\left(4y^2+4xy+x^2\right)-36\)

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

\(=\left(2y+x+6\right)\left(2y+x-6\right)\)

c) \(9x^2-12xy-25+4y^2\)

\(=\left(9x^2-12xy+4y^2\right)-25\)

\(=\left(3x-2y\right)^2-5^2\)

\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)

d) \(25x^2+10x-4y^2+1\)

\(=\left(25x^2+10x+1\right)-4y^2\)

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

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

\(x^4-5x^2y^2+4y^4\)

\(=\left(x^2\right)^2-2x^22y^2+\left(2y^2\right)^2-x^2y^2\)

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

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

minh hiền làm sai rồi

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

a,  x²+2xy+y²+2x+2y-15

<=> (x+y )²+2(x+y)+1-16

Đặt x+y =a

<=> a²+2a+1-4²

<=> (a+1)²-4²

<=> (a+5)(a-3) =>( x+y+5)(x+y-3)

b, x²-4xy+4y²-2x-4y-35

<=> (x-2y)²-2(x-2y)+1-36

Đặt (x-2y)  =b 

=> b²-2b+1-6²

<=> (b-1)²-6²

<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)

c, 

a,A= x^2+2xy+y^2+2x+2y-15

= (x+y)^2+(x+y)-15

Đặt x+y=a, ta có:

A=a^2+2a-15

  =a^2+2a+1-16

  =(a+1)^2-4^2

  =(a+1+4)(a+1-4)

  =(a+5)(a-3)

Thay a=x+y, ta có: A=(x+y+5)(x+y-3).

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