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3. 计算:
(1)$\sqrt{\frac{1}{3}}+\sqrt{12}=$
(2)$5\sqrt{ab}-3\sqrt{ab}=$
(1)$\sqrt{\frac{1}{3}}+\sqrt{12}=$
$\frac{7}{3}\sqrt{3}$
；(2)$5\sqrt{ab}-3\sqrt{ab}=$
$2\sqrt{ab}$
.答案
3. (1)$\frac{7}{3}\sqrt{3}$ (2)$2\sqrt{ab}$
解析
(1) $\sqrt{\frac{1}{3}}+\sqrt{12}=\frac{\sqrt{3}}{3}+2\sqrt{3}=\frac{\sqrt{3}}{3}+\frac{6\sqrt{3}}{3}=\frac{7}{3}\sqrt{3}$；
(2) $5\sqrt{ab}-3\sqrt{ab}=2\sqrt{ab}$
(2) $5\sqrt{ab}-3\sqrt{ab}=2\sqrt{ab}$
4. 计算:
(1)$\sqrt{0.5}+\sqrt{32}-2\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{8}}-\sqrt{75}$；
(2)$\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-(\frac{1}{2}\sqrt{2.4}-\sqrt{3\frac{3}{4}})$.
(1)$\sqrt{0.5}+\sqrt{32}-2\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{8}}-\sqrt{75}$；
(2)$\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-(\frac{1}{2}\sqrt{2.4}-\sqrt{3\frac{3}{4}})$.
答案
(1) $\sqrt{0.5}+\sqrt{32}-2\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{8}}-\sqrt{75}$
$=\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{4}-5\sqrt{3}$
$=(\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-5\sqrt{3})$
$=(\frac{2\sqrt{2}}{4}+\frac{16\sqrt{2}}{4}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-\frac{15\sqrt{3}}{3})$
$=\frac{17\sqrt{2}}{4}-\frac{17\sqrt{3}}{3}$
(2) $\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-(\frac{1}{2}\sqrt{2.4}-\sqrt{3\frac{3}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}\sqrt{\frac{12}{5}}-\sqrt{\frac{15}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}×\frac{2\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{2}$
$=(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{5})+(\frac{\sqrt{15}}{3}+\frac{\sqrt{15}}{2})$
$=0+\frac{2\sqrt{15}}{6}+\frac{3\sqrt{15}}{6}$
$=\frac{5\sqrt{15}}{6}$
$=\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{4}-5\sqrt{3}$
$=(\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-5\sqrt{3})$
$=(\frac{2\sqrt{2}}{4}+\frac{16\sqrt{2}}{4}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-\frac{15\sqrt{3}}{3})$
$=\frac{17\sqrt{2}}{4}-\frac{17\sqrt{3}}{3}$
(2) $\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-(\frac{1}{2}\sqrt{2.4}-\sqrt{3\frac{3}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}\sqrt{\frac{12}{5}}-\sqrt{\frac{15}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}×\frac{2\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{2}$
$=(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{5})+(\frac{\sqrt{15}}{3}+\frac{\sqrt{15}}{2})$
$=0+\frac{2\sqrt{15}}{6}+\frac{3\sqrt{15}}{6}$
$=\frac{5\sqrt{15}}{6}$
解析
(1) $\sqrt{0.5}+\sqrt{32}-2\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{8}}-\sqrt{75}$
$=\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{4}-5\sqrt{3}$
$=(\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-5\sqrt{3})$
$=(\frac{2\sqrt{2}}{4}+\frac{16\sqrt{2}}{4}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-\frac{15\sqrt{3}}{3})$
$=\frac{17\sqrt{2}}{4}-\frac{17\sqrt{3}}{3}$
(2) $\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-(\frac{1}{2}\sqrt{2.4}-\sqrt{3\frac{3}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}\sqrt{\frac{12}{5}}-\sqrt{\frac{15}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}×\frac{2\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{2}$
$=(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{5})+(\frac{\sqrt{15}}{3}+\frac{\sqrt{15}}{2})$
$=0+\frac{2\sqrt{15}}{6}+\frac{3\sqrt{15}}{6}$
$=\frac{5\sqrt{15}}{6}$
$=\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{2\sqrt{3}}{3}-\frac{\sqrt{2}}{4}-5\sqrt{3}$
$=(\frac{\sqrt{2}}{2}+4\sqrt{2}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-5\sqrt{3})$
$=(\frac{2\sqrt{2}}{4}+\frac{16\sqrt{2}}{4}-\frac{\sqrt{2}}{4})+(-\frac{2\sqrt{3}}{3}-\frac{15\sqrt{3}}{3})$
$=\frac{17\sqrt{2}}{4}-\frac{17\sqrt{3}}{3}$
(2) $\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-(\frac{1}{2}\sqrt{2.4}-\sqrt{3\frac{3}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}\sqrt{\frac{12}{5}}-\sqrt{\frac{15}{4}})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{1}{2}×\frac{2\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{2})$
$=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{2}$
$=(\frac{\sqrt{15}}{5}-\frac{\sqrt{15}}{5})+(\frac{\sqrt{15}}{3}+\frac{\sqrt{15}}{2})$
$=0+\frac{2\sqrt{15}}{6}+\frac{3\sqrt{15}}{6}$
$=\frac{5\sqrt{15}}{6}$
1. 下列式子中,正确的是(
A.$\sqrt{2}+\sqrt{3}=\sqrt{5}$
B.$m\sqrt{x}-n\sqrt{x}=(m-n)\sqrt{x}$
C.$\sqrt{a^2+b^2}=a+b$
D.$(\sqrt{4}+\sqrt{6})÷ 2=\sqrt{2}+\sqrt{3}$
B
)A.$\sqrt{2}+\sqrt{3}=\sqrt{5}$
B.$m\sqrt{x}-n\sqrt{x}=(m-n)\sqrt{x}$
C.$\sqrt{a^2+b^2}=a+b$
D.$(\sqrt{4}+\sqrt{6})÷ 2=\sqrt{2}+\sqrt{3}$
答案
1. B
2. 若$a+\sqrt{8}=\sqrt{18}$,则表示实数$a$的点会落在数轴的(
A.段①上
B.段②上
C.段③上
D.段④上
B
)A.段①上
B.段②上
C.段③上
D.段④上
答案
2. B
解析
$a+\sqrt{8}=\sqrt{18}$
$a=\sqrt{18}-\sqrt{8}$
$a=3\sqrt{2}-2\sqrt{2}=\sqrt{2}$
$\sqrt{1}＜\sqrt{2}＜\sqrt{4}$，即$1＜\sqrt{2}＜2$
表示实数$a$的点落在段②上
B
$a=\sqrt{18}-\sqrt{8}$
$a=3\sqrt{2}-2\sqrt{2}=\sqrt{2}$
$\sqrt{1}＜\sqrt{2}＜\sqrt{4}$，即$1＜\sqrt{2}＜2$
表示实数$a$的点落在段②上
B
3. 若长方形的长为$\sqrt{32}\ \mathrm{cm}$,宽为$\sqrt{18}\ \mathrm{cm}$,则该长方形的周长为
$14\sqrt{2}$
$\mathrm{cm}$.答案
3. $14\sqrt{2}$
解析
长方形的周长为 $2×(\sqrt{32}+\sqrt{18})$，化简得 $2×(4\sqrt{2}+3\sqrt{2}) = 2×7\sqrt{2}=14\sqrt{2}\ \mathrm{cm}$。
4. 计算:
(1)$(\sqrt{3}+1)(\sqrt{3}-1)+\sqrt{24}-(\frac{1}{2})^0$；
(2)$(\sqrt{3}+\sqrt{2}+1)(\sqrt{3}-\sqrt{2}+1)$.
(1)$(\sqrt{3}+1)(\sqrt{3}-1)+\sqrt{24}-(\frac{1}{2})^0$；
(2)$(\sqrt{3}+\sqrt{2}+1)(\sqrt{3}-\sqrt{2}+1)$.
答案
(1)
$\begin{aligned}&(\sqrt{3}+1)(\sqrt{3}-1)+\sqrt{24}-(\frac{1}{2})^0\\=&(\sqrt{3})^2 - 1^2 + 2\sqrt{6} - 1\\=&3 - 1 + 2\sqrt{6} - 1\\=&1 + 2\sqrt{6}\end{aligned}$
(2)
$\begin{aligned}&(\sqrt{3}+\sqrt{2}+1)(\sqrt{3}-\sqrt{2}+1)\\=&[(\sqrt{3}+1)+\sqrt{2}][(\sqrt{3}+1)-\sqrt{2}]\\=&(\sqrt{3}+1)^2 - (\sqrt{2})^2\\=&(3 + 2\sqrt{3} + 1) - 2\\=&4 + 2\sqrt{3} - 2\\=&2 + 2\sqrt{3}\end{aligned}$
1
$\begin{aligned}&(\sqrt{3}+1)(\sqrt{3}-1)+\sqrt{24}-(\frac{1}{2})^0\\=&(\sqrt{3})^2 - 1^2 + 2\sqrt{6} - 1\\=&3 - 1 + 2\sqrt{6} - 1\\=&1 + 2\sqrt{6}\end{aligned}$
(2)
$\begin{aligned}&(\sqrt{3}+\sqrt{2}+1)(\sqrt{3}-\sqrt{2}+1)\\=&[(\sqrt{3}+1)+\sqrt{2}][(\sqrt{3}+1)-\sqrt{2}]\\=&(\sqrt{3}+1)^2 - (\sqrt{2})^2\\=&(3 + 2\sqrt{3} + 1) - 2\\=&4 + 2\sqrt{3} - 2\\=&2 + 2\sqrt{3}\end{aligned}$
1
解析
(1)
$\begin{aligned}&(\sqrt{3}+1)(\sqrt{3}-1)+\sqrt{24}-(\frac{1}{2})^0\\=&(\sqrt{3})^2 - 1^2 + 2\sqrt{6} - 1\\=&3 - 1 + 2\sqrt{6} - 1\\=&1 + 2\sqrt{6}\end{aligned}$
(2)
$\begin{aligned}&(\sqrt{3}+\sqrt{2}+1)(\sqrt{3}-\sqrt{2}+1)\\=&[(\sqrt{3}+1)+\sqrt{2}][(\sqrt{3}+1)-\sqrt{2}]\\=&(\sqrt{3}+1)^2 - (\sqrt{2})^2\\=&(3 + 2\sqrt{3} + 1) - 2\\=&4 + 2\sqrt{3} - 2\\=&2 + 2\sqrt{3}\end{aligned}$
1
$\begin{aligned}&(\sqrt{3}+1)(\sqrt{3}-1)+\sqrt{24}-(\frac{1}{2})^0\\=&(\sqrt{3})^2 - 1^2 + 2\sqrt{6} - 1\\=&3 - 1 + 2\sqrt{6} - 1\\=&1 + 2\sqrt{6}\end{aligned}$
(2)
$\begin{aligned}&(\sqrt{3}+\sqrt{2}+1)(\sqrt{3}-\sqrt{2}+1)\\=&[(\sqrt{3}+1)+\sqrt{2}][(\sqrt{3}+1)-\sqrt{2}]\\=&(\sqrt{3}+1)^2 - (\sqrt{2})^2\\=&(3 + 2\sqrt{3} + 1) - 2\\=&4 + 2\sqrt{3} - 2\\=&2 + 2\sqrt{3}\end{aligned}$
1
5. 已知:$x=\sqrt{7}+\sqrt{5},y=\sqrt{7}-\sqrt{5}$.求下列各式的值.
(1)$x^2-xy+y^2$；
(2)$\frac{x}{y}-\frac{y}{x}$.
(1)$x^2-xy+y^2$；
(2)$\frac{x}{y}-\frac{y}{x}$.
答案
5. 解:(1)$\because x=\sqrt{7}+\sqrt{5}$,$y=\sqrt{7}-\sqrt{5}$,$\therefore x + y = (\sqrt{7}+\sqrt{5})+(\sqrt{7}-\sqrt{5}) = 2\sqrt{7}$,$x - y = (\sqrt{7}+\sqrt{5})-(\sqrt{7}-\sqrt{5}) = 2\sqrt{5}$,$xy = (\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5}) = 7 - 5 = 2$,$\therefore x^2 - xy + y^2 = (x + y)^2 - 3xy = 28 - 6 = 22$. (2)$\frac{x}{y}-\frac{y}{x}=\frac{x^2 - y^2}{xy}=\frac{(x + y)(x - y)}{xy}=\frac{2\sqrt{7}× 2\sqrt{5}}{2}=2\sqrt{35}$.
