2.用适当的符号填空:-|||-(1)a __ a,b,c ;;-|||-(2)0 __ x|{x)^2=0} ;-|||-(3)times __ xin R|{x)^2+1=0} ;-|||-(4) 0,1 __ N;-|||-(5)(0) __ x|{x)^2=x} ;-|||-(6)(2,1) __ x|{x)^2-3x+2=0} .

【答案】

$\left(1\right)\in $;$\left(2\right)\in $;

$\left(3\right)=$;$\left(4\right)\subsetneqq $;

$\left(5\right)\subsetneqq $;$\left(6\right)=$

【解析】

$\left(1\right)\because a$是$\{a,b,c\}$中的元素,

$\therefore a\in \left\{a,b,c\right\}$;

$\left(2\right)\because $由${x}^{2}=0$,解得$x=0$,

$\therefore \{x\mid {x}^{2}=0\}=\left\{0\right\}$,

$\therefore 0\in \{x\mid {x}^{2}=0\}$;

$\left(3\right)\because {x}^{2}+1=0$无实数根,

$\therefore \varnothing =\left\{x\in R\mid {x}^{2}+1=0\right\}$;

$\left(4\right)\because 0\in N$,$1\in N$,$2\in N$

$\therefore \left\{0,1\right\}\subsetneqq N$;

$\left(5\right)\because $由${x}^{2}=x$,解得$x=0$或$x=1$,

$\therefore \left\{x\mid {x}^{2}=x\right\}=\left\{0,1\right\}$,

$\therefore \left\{0\right\}\subsetneqq \left\{x\mid {x}^{2}=x\right\}$;

$\left(6\right)\because $由${x}^{2}-3x+2=0$,解得$x=2$或$x=1$,

$\therefore \left\{x\mid {x}^{2}-3x+2=0\right\}=\left\{2,1\right\}$,

$\therefore \left\{2,1\right\}=\left\{x\mid {x}^{2}-3x+2=0\right\}$.

故答案为:$\in $;$\in $;$=$;$\subsetneqq $;$\subsetneqq $;$=$.