设x 1 2 4-|||-f(x)= 2 2 1 2-x-|||-1 2 一 x 2 4-|||-1 x x+3 x+6，证明：x 1 2 4-|||-f(x)= 2 2 1 2-x-|||-1 2 一 x 2 4-|||-1 x x+3 x+6

(x)_(1)+2(x)_(2)+2(x)_(3)geqslant 4-|||-(x)_(1)+4(x)_(2)leqslant 20-|||-(x)_(1)+

((X)_(1)+(X)_(2)+(X)_(3))/3 b.X1-|||-c. ((X)_(1)+(X)_(2))/2 d. (2(X)_(1)+(X)_(2)

10.求下列函数的极值:-|||-(1) (x)=|x((x)^2-1)|;-|||-(2) (x)=dfrac (x({x)^2+1)}({x)^4-(x)^

解方程_(2)-3(x)_(3)+4(x)_(4)=-5-|||-x1- 2x3+324 =-4-|||-(x)_(1)+2(x)_(2)-5(x)_(4)=1

) (x)_(1)+4(x)_(2)-2(x)_(3)+8(x)_(4)=2 -(x)_(1)+2(x)_(2)+3(x)_(3)+4(x)_(4)=1 (x)

线性方程组 ) (x)_(1)+2(x)_(2)-2(x)_(3)=1 2(x)_(1)+4(x)_(2)-4(x)_(3)=2 3(x)_(1)+6(x)_

) 4(x)_(1)+6(x)_(2)geqslant 6 3(x)_(1)+2(x)_(2)geqslant 4 (x)_(1),(x)_(2)geqslan

例4 判断二次型-|||-((x)_(1),(x)_(2),(x)_(3))=(x)_(1)^2+2(x)_(2)^2+4(x)_(3)^2+2(x)_(1)(

4.设函数f(x,y)可导,且 f(1,-1)=-1 _(1)(1,-1)=2 _(2)(1,-1)=3, 又 F(x)=-|||-f[x^2,f(x^2,x

已知随机变量X的概率分布为-|||-X 0 1 2 3 4-|||-P 1/3 1/6 1/6 1/12 1/4-|||-求E(X),D(X), (-2x+1)