【例14】设a_(1)+a_(2)+...+a_(n)=0,求证:方程na_(n)x^n-1+(n-1)a_(n-1)x^n-2+...+2a_(2)x+a_(
+dfrac ({a)_(n)}(n+1)=0, 证明方程 _(0)+(a)_(1)x+-|||-_(2)(x)^2+... +(a)_(n)(x)^n=0 在
+(a)_(n-1)x=0 有一个正根 =(x)_(0) ,证明方程 _(0)n(x)^n-1+(a)_(1)(n-1)(x)^n-2+... +(a)_(n-
+(a)_(n-1)x=0 有一个正根 =(x)_(0), 证明方程 _(0)n(x)^n-1+(a)_(1)(n-1)(x)^n-2+... +(a)_(n-
若方程a_0x^n+a_1x^n-1+…+a_(n-1)x=0有一个正根x=x_0, 证明方程a_0nx^n-1+a_1(n-1)x^n-2+…+a_(n-1)
,-|||-;-|||-(3) (x)_(1)+(n-1)(x)_(2)+... +2(x)_(n-1)+(x)_(n)=0
设x_(0)=0,x_(n)=(1+2x_(n-1))/(1+x_(n-1))(n=1,2,3,...),则lim_(ntoinfty)x_(n)=设$x_{0
+dfrac ({a)_(n)}(n+1)=0, 证明多项式-|||-(x)=(a)_(0)+(a)_(1)x+... +(a)_(n)(x)^n-|||-在(
n-1 n-|||-1 2 ... n-1 0-|||-::-|||-1 2 ... 0 0-|||- ... 0 0;;
+(a)_(2n)x+(a)_(2n+1)=0-|||-至少有一个实根,其中a0,a1,···, _(2n+1) 均为常数, in N.-|||-5.证明:方程