a)\(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\)Giá trị của biểu thức trên tại \(a=4;b=-5;c=6\) là:
\(\dfrac{4-5-6}{4-\left(-5\right)+6}=-\dfrac{7}{15}\)
b: \(=\dfrac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\dfrac{2x-5y}{x-3y}\)
Đặt x/10=y/3=k
=>x=10k; y=3k
\(A=\dfrac{2\cdot10k-5\cdot3k}{10k-3\cdot3k}=\dfrac{5k}{k}=5\)
c: \(C=\left(\dfrac{x^3-y^3-x^3-y^3}{\left(x+y\right)\left(x-y\right)}\right):\dfrac{x^2-y^2-x^2}{x+y}\)
\(=\dfrac{-2y^3}{\left(x+y\right)\left(x-y\right)}\cdot\dfrac{x+y}{-y^2}=\dfrac{2y}{x-y}\)
\(=\dfrac{20}{9-10}=-20\)