a)  \(\frac{1}{9}x^4-2x^2y+9y^2=\left(\frac{1}{3}\right)^2\left(x^2\right)^2-2x^2y+\left(3y\right)^2\)

\(=\left(\frac{1}{3}x^2\right)^2-2\frac{1}{3}x^23y+\left(3y\right)^2\)

\(=\left(\frac{1}{3}x^2-3y\right)^2\)

b)  \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)

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

\(\frac{1}{9}x^4-2x^2y+9y^2\)

\(=\left(\frac{1}{3}x^2\right)^2-2\times\frac{1}{3}x^2\times3y+\left(3y\right)^2\)

\(=\left(\frac{1}{3}x^2-3y\right)^2\)

\(25x^2-20xy+4y^2\)

\(=\left(5x\right)^2-2\times5x\times2y+\left(2y\right)^2\)

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

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

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

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

2,\(=\left(x^2+2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

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

     \(=\left(x+y+z-t\right)\left(x+y-z+t\right)\)

3,\(=9x\left(x-y\right)-7\left(x-y\right)\)

    \(=\left(x-y\right)\left(9x-7\right)\)

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

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

a             \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)

b             \(4y^2+y+\frac{1}{16}=\left(2y\right)^2+2.2y.\frac{1}{4}+\left(\frac{1}{4}\right)^2=\left(2y+\frac{1}{4}\right)^2\)

\(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

bài 2 áp dụng hằng đẳng thức bạn nhé

bài 3\(A=\left(x^3+3x^2+3x+1\right)+5\)

              \(=\left(x+1\right)^3+5\)      thay x=19 vào ta được 

\(A=20^3+5=8005\)

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

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

thay x=11 vào ta được

\(B=\left(11-1\right)^3+1=10^3+1=1001\)

CÁC BẠN GIÚP MÌNH NHA MÌNH ĐANG CẦN GẤP 

TKS