... +({X)_(n)}^2)-|||-;(5) (mu )^2+dfrac (1)(3)((X)_(1)+(X)_(2)+(X)_(3))-|||-;(6) dfrac (1)({sigma )^2}sum _(i=1)^n({X)_(i)}^2 。

设sim N(mu ,1), X1、X2、X来自总体样本, hat (mu )=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+

【题文】已知(x+dfrac (1)(x))=(x)^2+dfrac (1)({x)^2}，求(x+dfrac (1)(x))=(x)^2+dfrac (1)(

已知(x+dfrac (1)(x))=(x)^2+dfrac (1)({x)^2}-3，求f(x)已知，求f(x)

x1，x2是取自总体N(μ，1)(μ未知)的样本。hat (mu )_(1)=dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) ;

4.设总体 sim N(mu ,(sigma )^2), x1,x2,x3为来自X的样本 hat (mu )=dfrac (1)(4)(x)_(1)+b(x)_

已知实数x满足(x)^2+dfrac (1)({x)^2}-3x-dfrac (3)(x)+2=0，求(x)^3+dfrac (1)({x)^3}的值.已知实数

dfrac (1)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+dfrac (2)(5)(X)_(3)-|||-D .dfrac (1)(7)(

设 (x+dfrac (1)(x))=(x)^2+dfrac (1)({x)^2}, 则 lim _(xarrow 3)f(x)= __

设x是非零实数,则 ^3+dfrac (1)({x)^3}=18-|||-__-|||-(1) +dfrac (1)(x)=3-|||-(2) ^2+dfrac

(D) (hat {mu )}_(4)=dfrac (1)(3)(X)_(1)+dfrac (3)(4)(X)_(2)-dfrac (1)(12)(X)_(3)