\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}\right)^2-2.\dfrac{1}{3}.y^4+\left(y^4\right)^2=\left(\dfrac{1}{3}+y^4\right)^2\)

\(y^8-\dfrac{2}{3}y^4+\dfrac{1}{9}=\left(y^4-\dfrac{1}{3}\right)^2\)

1: \(4x^2+4x+1=\left(2x+1\right)^2\)

2: \(x^2-20x+100=\left(x-10\right)^2\)

3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)

4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

a) \(24x^2-4xy\)

\(=4x\left(6x-y\right)\)

b) \(5x^3-10x^2+5x-20xy^2\)

\(=5x\left(x^2-10x+5-20y^2\right)\)

nhầm câu b nha 

\(5x^3-10x^2+5x-20xy^2\)

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

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

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

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

     y^2 - ( x^2 - 2x + 1 )

⇔ y^2 - ( x - 1 )^2

⇔ ( y - x - 1 ) ( y + x - 1 )

nha bạn 

  ( x + y )² - 2( x + y ) + 1 = 

= ( x + y - 1 )² ( áp dụng hằng đẳng thức thứ bình phương 1 hiệu nha )

Hok tốt!!!!!!!!!!!!!!!!

(x+y)²-2(x+y)+1

=(x+y)²-2.(x+y).1+1²

=(x+y-1)²

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

\(\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)