\(a,\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7x+7}{3x}\)
\(b,\dfrac{2a^2-2ab}{ac+ad-bc-bd}=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(a-b\right)\left(c+d\right)}=\dfrac{2a}{c+d}\)
\(c,\dfrac{x^2-xy}{y^2-x^2}=\dfrac{x\left(x-y\right)}{-\left(x-y\right)\left(x+y\right)}=\dfrac{x}{-x-y}\)
a) \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
b) \(\dfrac{2a^2-2ab}{ac+ad-bc-bd}=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}\) ( có sửa đề )
\(=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)
c) \(\dfrac{x^2-xy}{y^2-x^2}=\dfrac{-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{-x}{\left(x+y\right)}\)