Lời giải:
\(\begin{array}{l}7/9x^3-23x^2y+16xy^2\\=x(9x^2-12xy+16y^2)\\=x(3x-4y)^2\\8/9x^2+xy+\dfrac{1}{36}y^2\\=(3x)^2+2.3x.\dfrac{1}{6}y+\left(\dfrac{1}{6}y\right)^2\\=\left(3x+\dfrac{1}{6}y\right)^2\\9)16a^2-72a+81\\=(4a)^2-2.4a.9+9^2\\=(4a-9)^2\\10)x^2-36-x+6\\=(x-6)(x+6)-(x-6)\\=(x-6)(x+6-1)=(x-6)(x+5)\end{array}\)
