设 overrightarrow(alpha) = ((1)/(2) ), A = ( (1)/(sqrt(2)) & -(1)/(sqrt(2)) (1)/(sqrt(2)) & (1)/(sqrt(2)) 的长度为 ().

已知 alpha_1 = ((1)/(sqrt(3)), (1)/(sqrt(3)), (1)/(sqrt(3)) ), alpha_2 = (-(1)/

【例11】证明：当x>0时， 1+xln(x+sqrt(1+x^2))>sqrt(1+x^2).【例11】证明：当x>0时，$ 1+x\ln(x+\sqrt{1

练习3、设a∈(0,1),b∈(0,1),求证:sqrt(a^2)+b^(2)+sqrt((1-a)^2)+b^(2)+sqrt((1-a)^2)+(1-b)^

lim_(ntoinfty)((1)/(sqrt(n^2)+1)+(1)/(sqrt(n^2)+2)+...+(1)/(sqrt(n^2)+n))=_.$\li

12.证明lim_(ntoinfty)((1)/(sqrt(n^2)+1)+(1)/(sqrt(n^2)+2)+...+(1)/(sqrt(n^2)+n))=1

设=sqrt ({x)^2+(y)^2+(z)^2} 则|div(grad)|(1,0,1)= () .-|||-(A) -sqrt (2) (B) sqrt

设=dfrac (arcsin x)(sqrt {1-{x)^2}}（1）证明：=dfrac (arcsin x)(sqrt {1-{x)^2}}（2）求=df

(6) int_((1)/(sqrt(2)))^1(sqrt(1-x^2))/(x^2)dx;(6) $\int_{\frac{1}{\sqrt{2}}}^{1

lim _(xarrow 0)dfrac (sqrt {1+x)+sqrt (1-x)-2}(sqrt {1+{x)^2}-1}

A.sqrt(1+x) B.(sqrt(1+x))/(2) C.(sqrt(1+x))/(sqrt(x)) D.(sqrt(1+x))/(2sqrt(x)