4.欧氏空间R^3中的标准正交基是 () .-|||-(A) (dfrac (1)(sqrt {2)},0,dfrac (1)(sqrt {2)}), (dfrac (1)(sqrt {2)},0,-dfrac (1)(sqrt {2)}) ,(0,1,0)-|||-(B) (dfrac (1)(2),dfrac (1)(2),0), (-dfrac (1)(2),dfrac (1)(2),0), (0,0,1)-|||-(C) (dfrac (1)(sqrt {3)},dfrac (1)(sqrt {3)},dfrac (1)(sqrt {3)}), (dfrac (1)(sqrt {3)},-dfrac (1)(sqrt {3)},dfrac (1)(sqrt {3)}) ,(0,0,0)-|||-(D) (1,-1,1) ;(-1,1,1) ;(1,1,-1)

dfrac (1)(2)-|||-C .dfrac (sqrt {2)}(2)-|||-D .dfrac (sqrt {3)}(2)

dfrac ({mu )_(0)I}(pi R)(dfrac (1)(2)+dfrac (pi )(6))-|||-dfrac ({mu )_(0)I}(pi

数值求积公式(int )_(-1)^1f(x)dxapprox dfrac (2)(3)[ f(-dfrac (1)(sqrt {2)})+f(0)+f(dfr

已知氢原子的 (varphi )_(2{p)_(2)}=dfrac (1)(4sqrt {2pi {{a)_(0)}^3}}(dfrac (r)({a)_(0)

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

lim _(xarrow 4)dfrac (sqrt {2x+1)-3}(sqrt {x-2)-sqrt (2)}=..

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

已知复数=dfrac (2-2i)(1+sqrt {3)i}，则=dfrac (2-2i)(1+sqrt {3)i}=dfrac (2-2i)(1+sqrt {

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d