a: \(x^3-y^3=98\)
nên \(x-y=\dfrac{98}{x^2+xy+y^2}=2\)
mà x+y=8
nên x=(8+2)/2=5
=>y=3
b: \(x+y=\dfrac{x^3+y^3}{x^2-xy+y^2}=7\)
mà x-y=3
nên x=(7+3)/2=5
=>y=2
a,\(\dfrac{x^3-y^3}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^2+xy+y^2}=x-y=\dfrac{98}{49}=2\)
Ta có : \(\left\{{}\begin{matrix}x+y=8\\x-y=2\end{matrix}\right.< =>\left\{{}\begin{matrix}2x=10\\x+y=8\end{matrix}\right.< =>\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)
b, \(\dfrac{x^3+y^3}{x^2-xy+y^2}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2-xy+y^2}=x+y=\dfrac{133}{19}=7\)
Ta có : \(\left\{{}\begin{matrix}x+y=7\\x-y=3\end{matrix}\right.< =>\left\{{}\begin{matrix}2x=10\\x-y=3\end{matrix}\right.< =>\left\{{}\begin{matrix}x=5\\y=2\end{matrix}\right.\)
