\(a,=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\\ =\dfrac{4-5-6}{4+5+6}=-\dfrac{7}{15}\\ b,=\dfrac{8x\left(2x-5y\right)}{4x\left(2x-6y\right)}=\dfrac{4x-10y}{2x-6y}\left(1\right)\)
Lại có \(\dfrac{x}{y}=\dfrac{10}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{3}=\dfrac{4x-10y}{40-30}=\dfrac{4x-10y}{10}=\dfrac{2x-6y}{20-18}=\dfrac{2x-6y}{2}\)
Do đó \(\left(1\right)=\dfrac{10}{2}=5\)
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