a) \(-y^2+\dfrac{1}{9}\)
\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)
\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)
b) \(4^4-256\)
\(=4^4-4^4\)
\(=0\)
c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)
\(=\left(3x-9\right)^2-\left(2x+2\right)^2\)
\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)
\(=\left(5x-7\right)\left(x-11\right)\)
\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)
c. 9(x - 3)2 - 4(x + 1)2
= 32(x - 3)2 - 22(x + 1)2
= \(\left[3\left(x-3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
= (3x - 9)2 - (2x + 2)2
= (3x - 9 - 2x - 2)(3x - 9 + 2x + 2)
= (x - 11)(5x - 7)
d. 25x2 - \(\dfrac{1}{81}x^2y^2\)
= (5x)2 - \(\left(\dfrac{1}{9}xy\right)^2\)
= \(\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)\)
