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7. 能使等式$\sqrt{\dfrac{a}{a+3}}=\dfrac{\sqrt{a}}{\sqrt{a+3}}$成立的$a$的取值范围是
$a≥0$
.答案
7. $a≥0$
8. 设$a＞0$，$b＞0$，则下列运算错误的是 (
A.$\sqrt{ab}=\sqrt{a}·\sqrt{b}$
B.$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$
C.$(\sqrt{a})^{2}=a$
D.$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
B
)A.$\sqrt{ab}=\sqrt{a}·\sqrt{b}$
B.$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$
C.$(\sqrt{a})^{2}=a$
D.$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
答案
8. B
9. 计算：$\sqrt{\dfrac{3^{2}}{2^{5}}}$的值为 (
A.$\dfrac{3}{2}\sqrt{2}$
B.$\dfrac{3}{8}\sqrt{2}$
C.$\dfrac{3}{8}\sqrt{6}$
D.$8\sqrt{6}$
B
)A.$\dfrac{3}{2}\sqrt{2}$
B.$\dfrac{3}{8}\sqrt{2}$
C.$\dfrac{3}{8}\sqrt{6}$
D.$8\sqrt{6}$
答案
9. B
10. 若使代数式$\sqrt{x^{3}-x^{2}}=x\sqrt{x-1}$成立，则$x$的取值范围是 (
A.$x≥0$
B.$x＞0$
C.$x≥1$
D.$x＞1$
C
)A.$x≥0$
B.$x＞0$
C.$x≥1$
D.$x＞1$
答案
10. C
11. 已知$xy＞0$，化简二次根式$x\sqrt{-\dfrac{y}{x^{2}}}$的结果是 (
A.$\sqrt{y}$
B.$\sqrt{-y}$
C.$-\sqrt{y}$
D.$-\sqrt{-y}$
D
)A.$\sqrt{y}$
B.$\sqrt{-y}$
C.$-\sqrt{y}$
D.$-\sqrt{-y}$
答案
11. D
12. 化简：
(1)$\sqrt{169×121×4}$.
(2)$\sqrt{(2.5×10^{3})(1.6×10^{2})}$.
(3)$\sqrt{\dfrac{16}{125}}$.
(4)$\sqrt{1\dfrac{1}{80}}$.
(1)$\sqrt{169×121×4}$.
(2)$\sqrt{(2.5×10^{3})(1.6×10^{2})}$.
(3)$\sqrt{\dfrac{16}{125}}$.
(4)$\sqrt{1\dfrac{1}{80}}$.
答案
12. (1)286 (2)$200\sqrt{10}$ (3)$\frac{4}{25}\sqrt{5}$ (4)$\frac{9}{20}\sqrt{5}$
