dfrac (1)(2)x+dfrac (1)(3)=dfrac (1)(4)x-dfrac (1)(5)
(B) dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)-|||-(C) dfrac (1)(2)(X)_(1)+dfrac (1
解方程。 dfrac (1)(2)x-5= dfrac (1)(2) 5y-2y=18 dfrac (1)(2)+x= dfrac (1)
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)+
... +({X)_(n)}^2)-|||-;(5) (mu )^2+dfrac (1)(3)((X)_(1)+(X)_(2)+(X)_(3))-|||-;(6
dfrac (1)(x) C. -dfrac (1)({x)^2} D. dfrac (2)({x)^3}
[题目]已知 (dfrac (1)(x))=dfrac (5)(x)+2(x)^2, 则 f(x)=
lim _(xarrow 1)(dfrac (a)(1-{x)^2}+dfrac (x)(x-1))=dfrac (3)(2) = __
int dfrac (5x-1)({x)^2-x-2}dx= (.A.int dfrac (5x-1)({x)^2-x-2}dx= (B.int dfrac (
函数(x)=dfrac (1)(x-2)的间断点是(x)=dfrac (1)(x-2)(x)=dfrac (1)(x-2)、(x)=dfrac (1)(x-2)