\(a,=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=\dfrac{-\left(x+4\right)}{x}\\ b,=\dfrac{\left(x^2+x\right)+\left(3x+3\right)}{2\left(x+3\right)}=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{2\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
\(c,=\dfrac{5\left(x+y\right)^2.3.x.\left(x+y\right)}{5\left(x+y\right)^2.y}=\dfrac{3x\left(x+y\right)}{y}\\ d,=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{4}{5}\)
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