Answer: [tex]\bold{\bigg(\dfrac{1}{3}x+\dfrac{1}{5}y\bigg)^2}[/tex]
Step-by-step explanation:
The square of the trinomial exists only if (a + b)² = a² + 2ab + b²
[tex]\dfrac{1}{9}x^2+\dfrac{2}{15}xy+\dfrac{1}{25}y^2\\\\\\a=\sqrt{\dfrac{1}{9}x^2}=\dfrac{1}{3}x\\\\\\b=\sqrt{\dfrac{1}{25}y^2}=\dfrac{1}{5}y\\\\\\2\cdot a\cdot b=2\cdot \dfrac{1}{3}x\cdot \dfrac{1}{5}y = \dfrac{2}{15}xy\\\\\text{So, }\dfrac{1}{9}x^2+\dfrac{2}{15}xy+\dfrac{1}{25}y^2=\bigg(\dfrac{1}{3}x+\dfrac{1}{5}y\bigg)^2\\[/tex]
