(6) (x)=dfrac (sqrt {2x-{x)^2}}(lg (2x-1)) ;

函数=sqrt (16-{x)^2}+arcsin dfrac (2x-1)(7)的定义域为=sqrt (16-{x)^2}+arcsin dfrac (2x-

定积分(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0

求函数(x)=sqrt ({x)^2-x-6}+arcsin dfrac (2x-1)(7)的定义域.求函数的定义域.

[题目]求 lim _(xarrow 2)dfrac (sqrt {3x-2)-2}(sqrt {2x-1)-sqrt (3)}

=(e)^2x-|||-2. =(e)^-(x^2)-|||-3. =dfrac ({e)^x}(1+x)-|||-4. =dfrac (sqrt {(x+1)

求极限lim _(xarrow 1)dfrac (sqrt {2x-1)-1}({x)^2-1}求极限

=dfrac (x)(sqrt {1-{x)^2}}，则=dfrac (x)(sqrt {1-{x)^2}}=_________.，则=_________.

9.函数 =sqrt ({x)^2-x-6}+arcsin dfrac (2x-1)(7) 的定义域是 ()-|||-

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

(8)(int )_(1)^2sqrt (2x-{x)^2}dx=__________________.(8)__________________.