\(\sqrt{20}\)+\(\sqrt{80}\)-\(\sqrt{45}\)
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9.
\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)
\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=-\sqrt{5}\)
10.
\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)
\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)
\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)
\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)
Lời giải:
a.
$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$
b.
$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$
\(a,\sqrt{20}-\sqrt{45}+3\sqrt{80}\)
b, Tính:\(\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\sqrt{3}\)
a)\(\sqrt{20}-\sqrt{45}+3\sqrt{80}\)
= \(2\sqrt{5}-3\sqrt{5}+3.4\sqrt{5}\)
= ( 2 - 3 + 12 )\(\sqrt{5}\)
= 11\(\sqrt{5}\)
b) (\(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\) )\(\sqrt{3}\)
= 6 + 18 - 9
=15
\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
\(=-5\)
\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)
\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)
\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)
\(=\sqrt{5}.\left(5-12+6-4\right)\)
\(=-5\sqrt{5}\)
\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)
\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)
\(=2.3+5-5.2\)
\(=1\)
\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)
\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)
\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)
\(=0\)
\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)
\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)
\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)
\(=7\sqrt{3}\)
\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)
\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
L=2*3+5-5*2=5-4=1
N=8căn 2-8căn2-15căn3+15căn 3=0
O=10căn 3-6căn3+3căn3=7căn 3
\(2\sqrt{5}+4\sqrt{5}-3\sqrt{5}\) = \(3\sqrt{5}\)
\(=2\sqrt{5}+4\sqrt{5}-3\sqrt{5}=3\sqrt{5}\)