\(\left\{{}\begin{matrix}x+y=-3\\xy=-28\end{matrix}\right.\)
Nên \(\left(x+y\right)^2=9\)
\(x^2+2xy+y^2=9\)
\(\Rightarrow x^2-56+y^2=9\)
\(\Rightarrow x^2+y^2=65\)(1)
Ta có:
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=-3\left(65+28\right)=-3.93=-279\)(2)
\(x^4+y^4=x^4+y^4+2\left(xy\right)^2-2\left(xy\right)^2\)
\(=\left(x^2+y^2\right)^2-2\left(xy\right)^2=65^2-18=4207\)
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