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2015.2李建泉(数论)讲义

若关于$x,y$的多项式$x^{m-2}y^2+mx^{m-2}y+nx^3y^{m-3}-2x^{m-3}y+m+n$,化简后是四次三项式,求$m,n$的值.
\begin{solution}
\textbf{解法一.}算多元多项式的次数应该把$x$和$y$的次数加在一起,则$x^ {m-2}y^2$, $mx^{m-2}y$, $nx^3y^{m-3}$, $2x^{m-3}y$分别是$m,m-1,m,m-2$次.

(1)若$m=4$,则
$$\begin{align*} &x^{m-2}y^2+mx^{m-2}y+nx^3y^{m-3}-2x^{m-3}y+m+n \\ &=x^2y^2+4x^2y+nx^3y-2xy+4+n \end{align*}$$
不可能是三项式,矛盾.

(2)若$m=5$,则
$$\begin{align*} & x^{m-2}y^2+mx^{m-2}y+nx^3y^{m-3}-2x^{m-3}y+m+n \\ & =x^3y^2+5x^3y+nx^3y^2-2x^2y+5+n \\ &=\left( 1+n \right) x^3y^2+5x^3y-2x^2y+5+n, \end{align*}$$
当$n+1=0$,即$n=-1$时,它为四次三项式.

因此$m=5,n=-1$.

\textbf{解法二.}我们先让项数减少,显然$mx^{m-2}y$和$-2x^{m-3}y$不可能是同类项,故只有$x^{m-2}y^2$和$nx^3y^{m-3}$是同类项,所以$m-2=3,2=m-3$,即$m=5$,则
$$\begin{align*} & x^{m-2}y^2+mx^{m-2}y+nx^3y^{m-3}-2x^{m-3}y+m+n \\ & =\left( 1+n \right) x^3y^2+5x^3y-2x^2y+5+n \end{align*}$$
由于它是四次三项式,则$n+1=0$,则$n=-1$.

因此$m=5,n=-1$.
\end{solution}

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