\(a.\dfrac{x^2}{x+5}-\dfrac{25}{x+5}\\ \dfrac{x^2-25}{x+5}=\dfrac{\left(x-5\right)\left(x+5\right)}{x+5}=x-5\\ b.\dfrac{x^2}{x-y}+\dfrac{y^2}{y-x}\\ =\dfrac{x^2}{x-y}-\dfrac{y^2}{x-y}=\dfrac{x^2-y^2}{x-y}\\ =\dfrac{\left(x-y\right)\left(x+y\right)}{x-y}=x+y\)
\(c.\dfrac{3x+y^2}{x^2y}-\dfrac{3y-x^2}{xy^2}\\ =\dfrac{y\left(3x+y^2\right)}{x^2y^2}-\dfrac{x\left(3x-y^2\right)}{x^2y^2}\\ =\dfrac{3xy+y^3}{x^2y^2}-\dfrac{3x^2-xy^2}{x^2y^2}\\ =\dfrac{3xy+y^3-3x^2+xy^2}{x^2y^2}\)
\(d.\dfrac{2}{5x^2y}+\dfrac{4}{15xy^2}-\dfrac{2}{3xy}\\ =\dfrac{3y\cdot2}{15x^2y^2}+\dfrac{4x}{15x^2y^2}-\dfrac{10xy}{15x^2y^2}\\ =\dfrac{6y+4x-10xy}{15x^2y^2}\)
e: \(\dfrac{b}{2a^2-ab}+\dfrac{4a}{b^2-2ab}\)
\(=\dfrac{b}{a\left(2a-b\right)}-\dfrac{4a}{2ab-b^2}\)
\(=\dfrac{b}{a\left(2a-b\right)}-\dfrac{4a}{b\left(2a-b\right)}\)
\(=\dfrac{b^2-4a^2}{ab\left(2a-b\right)}=\dfrac{\left(b-2a\right)\left(b+2a\right)}{ab\left(2a-b\right)}\)
\(=\dfrac{-b-2a}{ab}\)
g: \(\dfrac{5}{2x+6}-\dfrac{3}{x^2+3x}\)
\(=\dfrac{5}{2\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{5x-6}{2x\left(x+3\right)}\)
h: \(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(2-x\right)\left(3-x\right)}+\dfrac{3}{\left(1-x\right)\left(x-3\right)}\)
\(=\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-2\right)\left(x-3\right)}-\dfrac{3}{\left(x-1\right)\left(x-3\right)}\)
\(=\dfrac{x-3+2\left(x-1\right)-3\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{3x-5-3x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
i: \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x+1\right)\left(x-1\right)}+\dfrac{1}{x-1}+1\)
\(=\dfrac{x^2\left(x-1\right)+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x^2+x^2+3x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3+3x}{\left(x-1\right)\left(x+1\right)}\)
k: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)}-\dfrac{1}{\left(b-c\right)\left(a-c\right)}-\dfrac{1}{\left(a-c\right)\left(a-b\right)}\)
\(=\dfrac{a-c-a+b-b+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}==0\)
