a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)
b) \(a^2-9=\left(a-3\right)\left(a+3\right)\)
c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)
d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)
e) \(\left(2a+b\right)^2-a^2=\left(2a+b-a\right)\left(2a+b+a\right)=\left(a+b\right)\left(3a+b\right)\)
f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)
a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$
b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$
c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$
d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$
e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$
f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$