登录
1.「2026河南洛阳汝阳期末，★☆」下列二次根式运算正确的是（
A.$\sqrt{4}+\sqrt{9}=\sqrt{13}$
B.$\sqrt{3}×\sqrt{12}=6$
C.$\sqrt{18}÷\sqrt{2}=9$
D.$\sqrt{5}-\sqrt{2}=\sqrt{3}$
B
）A.$\sqrt{4}+\sqrt{9}=\sqrt{13}$
B.$\sqrt{3}×\sqrt{12}=6$
C.$\sqrt{18}÷\sqrt{2}=9$
D.$\sqrt{5}-\sqrt{2}=\sqrt{3}$
答案
√4+√9=2+3=5,√3×√12=6,√18÷√2=√9=3,√5与-√2不能合并,故选项A,C,D运算错误,不符合题意,选项B运算正确,符合题意.故选B.
2.「★☆」下面是小明和小亮的计算过程,则选项中判断正确的是
(
小明:$\sqrt{3} × \sqrt{6} = \sqrt{3 × 6} = \sqrt{3 × 3 × 2} = \sqrt{3^2 × 2} = 3\sqrt{2}$.
小亮:$2\sqrt{5} × \sqrt{10} = 2\sqrt{5} × \sqrt{5 × 2} = 2\sqrt{5} × \sqrt{5} × \sqrt{2} = 2 × (\sqrt{5})^2 × \sqrt{2} = 10\sqrt{2}$.
A.只有小明的做法正确
B.两人的做法都不正确
C.小明在计算时用到了$\sqrt{a} · \sqrt{b} = \sqrt{ab} (a ≥ 0, b ≥ 0)$
D.小亮在计算时用到了$\sqrt{a^2} = a (a ≥ 0)$
(
C
)小明:$\sqrt{3} × \sqrt{6} = \sqrt{3 × 6} = \sqrt{3 × 3 × 2} = \sqrt{3^2 × 2} = 3\sqrt{2}$.
小亮:$2\sqrt{5} × \sqrt{10} = 2\sqrt{5} × \sqrt{5 × 2} = 2\sqrt{5} × \sqrt{5} × \sqrt{2} = 2 × (\sqrt{5})^2 × \sqrt{2} = 10\sqrt{2}$.
A.只有小明的做法正确
B.两人的做法都不正确
C.小明在计算时用到了$\sqrt{a} · \sqrt{b} = \sqrt{ab} (a ≥ 0, b ≥ 0)$
D.小亮在计算时用到了$\sqrt{a^2} = a (a ≥ 0)$
答案
小明:$\sqrt{3} × \sqrt{6} = \sqrt{3 × 6} = \sqrt{3 × 3 × 2} = \sqrt{3^2 × 2} = 3\sqrt{2}$,做法是正确的,计算时用到了$\sqrt{a} · \sqrt{b} = \sqrt{ab} (a ≥ 0, b ≥ 0)$.
小亮:$2\sqrt{5} × \sqrt{10} = 2\sqrt{5} × \sqrt{5 × 2} = 2\sqrt{5} × \sqrt{5} × \sqrt{2} = 2 × (\sqrt{5})^2 × \sqrt{2} = 10\sqrt{2}$,做法是正确的,计算时用到了$(\sqrt{a})^2 = a(a ≥ 0)$,故选项A,B,D不正确,选项C正确.故选C.
小亮:$2\sqrt{5} × \sqrt{10} = 2\sqrt{5} × \sqrt{5 × 2} = 2\sqrt{5} × \sqrt{5} × \sqrt{2} = 2 × (\sqrt{5})^2 × \sqrt{2} = 10\sqrt{2}$,做法是正确的,计算时用到了$(\sqrt{a})^2 = a(a ≥ 0)$,故选项A,B,D不正确,选项C正确.故选C.
3.「2026山西晋城期末，★☆☆」计算：$(\sqrt{3}+\sqrt{27}-\sqrt{12})÷\sqrt{6}=$
$\sqrt{2}$
.答案
答案 $\sqrt{2}$
解析 原式$=(\sqrt{3}+3\sqrt{3}-2\sqrt{3})÷\sqrt{6}=2\sqrt{3}÷\sqrt{6}=\sqrt{2}$.
解析 原式$=(\sqrt{3}+3\sqrt{3}-2\sqrt{3})÷\sqrt{6}=2\sqrt{3}÷\sqrt{6}=\sqrt{2}$.
4.「2026 河北秦皇岛期末，★☆」计算：
(1) $2\sqrt{12} - 6\sqrt{\dfrac{1}{3}} + 3\sqrt{48}$.
(2) $2\sqrt{\dfrac{1}{3}} × 3\sqrt{18} ÷ \sqrt{\dfrac{5}{6}}$.
(3) $\sqrt{40} ÷ \sqrt{5} + \sqrt{\dfrac{1}{4}} × \sqrt{72} - \sqrt{32}$.
(4) $\sqrt{2} × \sqrt{6} + |\sqrt{3} - 2| - (\sqrt{5} + 2)(\sqrt{5} - 2)$.
(1) $2\sqrt{12} - 6\sqrt{\dfrac{1}{3}} + 3\sqrt{48}$.
(2) $2\sqrt{\dfrac{1}{3}} × 3\sqrt{18} ÷ \sqrt{\dfrac{5}{6}}$.
(3) $\sqrt{40} ÷ \sqrt{5} + \sqrt{\dfrac{1}{4}} × \sqrt{72} - \sqrt{32}$.
(4) $\sqrt{2} × \sqrt{6} + |\sqrt{3} - 2| - (\sqrt{5} + 2)(\sqrt{5} - 2)$.
答案
(1) 原式$=4\sqrt{3}-2\sqrt{3}+12\sqrt{3}=14\sqrt{3}$.
(2) 原式$=6\sqrt{6}÷\sqrt{\dfrac{5}{6}}=6\sqrt{6×\dfrac{6}{5}}=\dfrac{36\sqrt{5}}{5}$.
(3) 原式$=\sqrt{8}+\sqrt{18}-4\sqrt{2}=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}=\sqrt{2}$.
(4) 原式$= \sqrt{12}+2-\sqrt{3}-(5-2\sqrt{5}+2\sqrt{5}-4)= 2\sqrt{3}+2-\sqrt{3}-1= \sqrt{3}+1$.
(2) 原式$=6\sqrt{6}÷\sqrt{\dfrac{5}{6}}=6\sqrt{6×\dfrac{6}{5}}=\dfrac{36\sqrt{5}}{5}$.
(3) 原式$=\sqrt{8}+\sqrt{18}-4\sqrt{2}=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}=\sqrt{2}$.
(4) 原式$= \sqrt{12}+2-\sqrt{3}-(5-2\sqrt{5}+2\sqrt{5}-4)= 2\sqrt{3}+2-\sqrt{3}-1= \sqrt{3}+1$.
