2. 化简。
(1)$\sqrt{16×81}$; (2)$\sqrt{\dfrac{19}{4}}×\sqrt{\dfrac{8}{19}}$;
(3)$\sqrt{3^{2}×5}$; (4)$3\sqrt{2}×2\sqrt{6}$。
(1)$\sqrt{16×81}$; (2)$\sqrt{\dfrac{19}{4}}×\sqrt{\dfrac{8}{19}}$;
(3)$\sqrt{3^{2}×5}$; (4)$3\sqrt{2}×2\sqrt{6}$。
答案
(1) $\sqrt{16×81}=\sqrt{16}×\sqrt{81}=4×9=36$;
(2) $\sqrt{\dfrac{19}{4}}×\sqrt{\dfrac{8}{19}}=\sqrt{\dfrac{19}{4}×\dfrac{8}{19}}=\sqrt{2}$;
(3) $\sqrt{3^{2}×5}=\sqrt{3^{2}}×\sqrt{5}=3\sqrt{5}$;
(4) $3\sqrt{2}×2\sqrt{6}=3×2×\sqrt{2×6}=6×\sqrt{12}=6×2\sqrt{3}=12\sqrt{3}$。
(2) $\sqrt{\dfrac{19}{4}}×\sqrt{\dfrac{8}{19}}=\sqrt{\dfrac{19}{4}×\dfrac{8}{19}}=\sqrt{2}$;
(3) $\sqrt{3^{2}×5}=\sqrt{3^{2}}×\sqrt{5}=3\sqrt{5}$;
(4) $3\sqrt{2}×2\sqrt{6}=3×2×\sqrt{2×6}=6×\sqrt{12}=6×2\sqrt{3}=12\sqrt{3}$。
3. 计算。
(1)$\sqrt{8}×\sqrt{2}$;
(2)$\sqrt{(\dfrac{1}{2})^{3}}×\sqrt{8}$;
(3)$3\sqrt{5}×2\sqrt{10}$;
(4)$\sqrt{300}$;
(5)$\sqrt{3x}·\sqrt{\dfrac{1}{3}xy}$;
(6)$\dfrac{1}{3}\sqrt{3x^{3}y^{2}}·\dfrac{1}{2}\sqrt{12xy^{2}}$。
(1)$\sqrt{8}×\sqrt{2}$;
(2)$\sqrt{(\dfrac{1}{2})^{3}}×\sqrt{8}$;
(3)$3\sqrt{5}×2\sqrt{10}$;
(4)$\sqrt{300}$;
(5)$\sqrt{3x}·\sqrt{\dfrac{1}{3}xy}$;
(6)$\dfrac{1}{3}\sqrt{3x^{3}y^{2}}·\dfrac{1}{2}\sqrt{12xy^{2}}$。
答案
(1)
$\sqrt{8}×\sqrt{2}=\sqrt{8×2}=\sqrt{16} = 4$
(2)
$\sqrt{(\frac{1}{2})^{3}}×\sqrt{8}=\sqrt{(\frac{1}{2})^{3}×8}=\sqrt{\frac{1}{8}×8}=\sqrt{1}=1$
(3)
$3\sqrt{5}×2\sqrt{10}=(3×2)×\sqrt{5×10}=6\sqrt{50}=6\sqrt{25×2}=6×5\sqrt{2}=30\sqrt{2}$
(4)
$\sqrt{300}=\sqrt{100×3}=\sqrt{100}×\sqrt{3}=10\sqrt{3}$
(5)
$\sqrt{3x}·\sqrt{\frac{1}{3}xy}=\sqrt{3x·\frac{1}{3}xy}=\sqrt{x^{2}y}=x\sqrt{y}$
(6)
$\frac{1}{3}\sqrt{3x^{3}y^{2}}·\frac{1}{2}\sqrt{12xy^{2}}=(\frac{1}{3}×\frac{1}{2})\sqrt{3x^{3}y^{2}·12xy^{2}}$
$=\frac{1}{6}\sqrt{36x^{4}y^{4}}=\frac{1}{6}×6x^{2}y^{2}=x^{2}y^{2}$
$\sqrt{8}×\sqrt{2}=\sqrt{8×2}=\sqrt{16} = 4$
(2)
$\sqrt{(\frac{1}{2})^{3}}×\sqrt{8}=\sqrt{(\frac{1}{2})^{3}×8}=\sqrt{\frac{1}{8}×8}=\sqrt{1}=1$
(3)
$3\sqrt{5}×2\sqrt{10}=(3×2)×\sqrt{5×10}=6\sqrt{50}=6\sqrt{25×2}=6×5\sqrt{2}=30\sqrt{2}$
(4)
$\sqrt{300}=\sqrt{100×3}=\sqrt{100}×\sqrt{3}=10\sqrt{3}$
(5)
$\sqrt{3x}·\sqrt{\frac{1}{3}xy}=\sqrt{3x·\frac{1}{3}xy}=\sqrt{x^{2}y}=x\sqrt{y}$
(6)
$\frac{1}{3}\sqrt{3x^{3}y^{2}}·\frac{1}{2}\sqrt{12xy^{2}}=(\frac{1}{3}×\frac{1}{2})\sqrt{3x^{3}y^{2}·12xy^{2}}$
$=\frac{1}{6}\sqrt{36x^{4}y^{4}}=\frac{1}{6}×6x^{2}y^{2}=x^{2}y^{2}$
1. 当$x< y$时,化简二次根式$\sqrt{-x^{3}y}$的正确结果是
$-x\sqrt{-xy}$
。答案
1. $-x\sqrt{-xy}$
2. 使$\sqrt{(x + 2)(3 - x)}=\sqrt{x + 2}·\sqrt{3 - x}$成立的条件是(
A.$x≤3$
B.$x≥ - 2$
C.$-2≤ x≤3$
D.$-2< x<3$
C
)A.$x≤3$
B.$x≥ - 2$
C.$-2≤ x≤3$
D.$-2< x<3$
答案
2. C
3. 若$\sqrt{50a}$是整数,则正整数$a$的最小值为
2
。答案
3. 2
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