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1. 若$\sqrt{x+1}$，$\sqrt{8-x}$都为整数，则$x$的值是
-1或8
.答案
-1或8 解析:根据题意,得$\begin{cases} x+1≥0, \\ 8-x≥0, \end{cases}$解得$-1 ≤ x ≤ 8$.
$\because \sqrt{x+1},\sqrt{8-x}$都为整数,$\therefore x=-1$或$8$.
$\because \sqrt{x+1},\sqrt{8-x}$都为整数,$\therefore x=-1$或$8$.
2. 已知$a,b,c$满足：$b=\sqrt{-(a-3)^2}+4$，$c$的平方根等于它本身，则$a+\sqrt{b-c}$的值为________。
答案
5 解析:$\because -(a-3)^2 ≥ 0, \therefore a=3$. 把$a=3$代入$b=\sqrt{-(a-3)^2}+4$得$b=4$.$\because c$的平方根等于它本身,$\therefore c=0$,$\therefore a+\sqrt{b-c}=3+\sqrt{4-0}=5$.
3. (1)若$x,y$均为有理数,且$\sqrt{x-1}+\sqrt{1-x}+2y-1=0$,求$\sqrt{15x+2y}$的平方根.
(2)若$y=\sqrt{x-3}+\sqrt{3-x}+4$,求$\sqrt{x^2-2xy+y^2}+\sqrt{4x^2-4xy+y^2}$的值.
(3)若$\sqrt{(x-1000)^2}+(\sqrt{998-x})^2=2000$,$y=\sqrt{m+8}+\sqrt{m-1}+\sqrt{1-m}$,求$y-x$的平方根.
(2)若$y=\sqrt{x-3}+\sqrt{3-x}+4$,求$\sqrt{x^2-2xy+y^2}+\sqrt{4x^2-4xy+y^2}$的值.
(3)若$\sqrt{(x-1000)^2}+(\sqrt{998-x})^2=2000$,$y=\sqrt{m+8}+\sqrt{m-1}+\sqrt{1-m}$,求$y-x$的平方根.
答案
(1)$\because \sqrt{x-1}+\sqrt{1-x}+2y-1=0$,$\therefore x-1≥0$,$1-x≥0$,解得$x=1$,$\therefore 2y-1=0$,$\therefore y=\dfrac{1}{2}$,$\therefore \sqrt{15x+2y}=\sqrt{15+1}=\sqrt{16}=4$,$\therefore \sqrt{15x+2y}$的平方根为$\pm2$.
(2)$\because y=\sqrt{x-3}+\sqrt{3-x}+4$,$\therefore x-3≥0$,$3-x≥0$,解得$x=3$,$y=4$,$\sqrt{x^2-2xy+y^2}+\sqrt{4x^2-4xy+y^2}=\sqrt{(x-y)^2}+\sqrt{(2x-y)^2}=1+2=3$.
(3)根据题意,得$998-x≥0$,解得$x≤998$,$\therefore \sqrt{(x-1\ 000)^2}=1\ 000-x$.等式可变为$1\ 000-x+998-x=2\ 000$,解得$x=-1$.根据题意,得$m-1≥0$且$1-m≥0$,解得$m≥1$且$m≤1$,$\therefore m=1$,$y=\sqrt{1+8}=3$,$\therefore y-x=3-(-1)=3+1=4$.$\because (\pm2)^2=4$,$\therefore 4$的平方根是$\pm2$,即$y-x$的平方根是$\pm2$.
(2)$\because y=\sqrt{x-3}+\sqrt{3-x}+4$,$\therefore x-3≥0$,$3-x≥0$,解得$x=3$,$y=4$,$\sqrt{x^2-2xy+y^2}+\sqrt{4x^2-4xy+y^2}=\sqrt{(x-y)^2}+\sqrt{(2x-y)^2}=1+2=3$.
(3)根据题意,得$998-x≥0$,解得$x≤998$,$\therefore \sqrt{(x-1\ 000)^2}=1\ 000-x$.等式可变为$1\ 000-x+998-x=2\ 000$,解得$x=-1$.根据题意,得$m-1≥0$且$1-m≥0$,解得$m≥1$且$m≤1$,$\therefore m=1$,$y=\sqrt{1+8}=3$,$\therefore y-x=3-(-1)=3+1=4$.$\because (\pm2)^2=4$,$\therefore 4$的平方根是$\pm2$,即$y-x$的平方根是$\pm2$.
4. 已知 $ a $ 满足 $ |2000 - a| + \sqrt{a - 2001} = a $，求 $ a - 2000^2 $ 的值.
答案
根据题意,得$a-2\ 001≥0$,$\therefore a≥2\ 001$,$\therefore |2\ 000-a|=a-2\ 000$.$\because |2\ 000-a|+\sqrt{a-2\ 001}=a$,$\therefore a-2\ 000+\sqrt{a-2\ 001}=a$,$\therefore \sqrt{a-2\ 001}=2\ 000$,$\therefore a-2\ 001=2\ 000^2$,$\therefore a-2\ 000^2=2\ 001$.
5. 如果关于$ x $的方程$ 2+\sqrt{4x+3}=k $无解，那么$ k $的取值范围是
k＜2
.答案
$k＜2$ 解析:$2+\sqrt{4x+3}=k$,$\sqrt{4x+3}=k-2$,$\because \sqrt{4x+3}$的结果是非负数,$\therefore $当$k-2＜0$,即$k＜2$时,方程$2+\sqrt{4x+3}=k$无解.
6. 已知$(x+y-1)^2$与$\sqrt{2x-y+4}$互为相反数，则$(\dfrac{1}{2}x)^{-y}$的值为________。
答案
4 解析:$\because (x+y-1)^2$与$\sqrt{2x-y+4}$互为相反数,$\therefore (x+y-1)^2+\sqrt{2x-y+4}=0$,由非负数的性质得$x+y-1=0$,$2x-y+4=0$,解得$x=-1$,$y=2$,$\therefore (\dfrac{1}{2}x)^{-y}=4$.
7. 已知$a,b$为有理数，且$\sqrt{1+a}-(b-1)\sqrt{1-b}=0$，则$a^{9999}-b^{10000}$的值为________.
答案
-2 解析:$\because \sqrt{1+a}-(b-1)\sqrt{1-b}=0$,$\therefore \sqrt{1+a}+(1-b)·\sqrt{1-b}=0$.$\because \sqrt{1+a}≥0$,$\sqrt{1-b}≥0$,$\therefore 1+a=0$,$1-b=0$,解得$a=-1$,$b=1$,$\therefore a^{999}-b^{1\ 000}=(-1)^{999}-1^{1\ 000}=-1-1=-2$.
8. (1)若$a,b,c$满足$\sqrt{2a+b-4}+|a-c+1|=\sqrt{b-c}+\sqrt{c-b}$，则$a+b+c$的平方根为________。
(2)若$a,b,c,d$满足$|a-4|+\sqrt{(2+b)^2}+\sqrt{c-5}+\sqrt{d-3}-\sqrt{3-d}=0$，则$a+b-c+d$的值为________。
(2)若$a,b,c,d$满足$|a-4|+\sqrt{(2+b)^2}+\sqrt{c-5}+\sqrt{d-3}-\sqrt{3-d}=0$，则$a+b-c+d$的值为________。
答案
(1)$\pm\sqrt{5}$ 解析:根据题意,得$b-c≥0$且$c-b≥0$,$\therefore b≥ c$且$c≥ b$,$\therefore b=c$,$\therefore$ 等式可变为$\sqrt{2a+b-4}+|a-b+1|=0$,由非负数的性质得$\begin{cases} 2a+b-4=0, \\ a-b+1=0, \end{cases}$解得$\begin{cases} a=1, \\ b=2, \end{cases}$$\therefore c=2$,$a+b+c=1+2+2=5$,$\therefore a+b+c$的平方根是$\pm\sqrt{5}$.
(2)$0$ 解析:根据题意,得$d-3≥0$,$3-d≥0$,$\therefore d=3$,$\therefore$ 等式可变为$|a-4|+\sqrt{(2+b)^2}+\sqrt{c-5}=0$,由非负数的性质得$a-4=0$,$2+b=0$,$c-5=0$,解得$a=4$,$b=-2$,$c=5$,$\therefore a+b-c+d=4-2-5+3=0$.
(2)$0$ 解析:根据题意,得$d-3≥0$,$3-d≥0$,$\therefore d=3$,$\therefore$ 等式可变为$|a-4|+\sqrt{(2+b)^2}+\sqrt{c-5}=0$,由非负数的性质得$a-4=0$,$2+b=0$,$c-5=0$,解得$a=4$,$b=-2$,$c=5$,$\therefore a+b-c+d=4-2-5+3=0$.
9. 已知 $a,b,c$ 满足 $|2a-4|+|b+2|+\sqrt{(a-3)b^2}+a^2+c^2=2+2ac$，求 $a-b+c$ 的值.
答案
$\because |2a-4|+|b+2|+\sqrt{(a-3)b^2}+a^2+c^2=2+2ac$,$\therefore |2a-4|+|b+2|+\sqrt{(a-3)b^2}+(a-c)^2=2$,由于左边各项的值都是大于等于0,$\therefore $当$b=0$时,则只能$2a-4=0$,$a-c=0$,即$a=c=2$,$b=0$,$\therefore a-b+c=4$;当$b\ne0$时,则$a≥3$,在$a≥3$的情况下,$|2a-4|≥2$,$\therefore a=3$且$c=a$,$b=-2$,$\therefore a-b+c=8$.综上所述,$a-b+c$的值为4或8.
