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a) \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
= \(\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
= \(-16\sqrt{3}\)
b) \(\left(a.\sqrt{\dfrac{a}{b}}+2\sqrt{ab}+b.\sqrt{\dfrac{b}{a}}\right)\sqrt{\dfrac{a}{b}}\)
= \(\dfrac{a^2}{b}+2a+b\) = \(\dfrac{a^2+\left(2a+b\right)b}{b}\) = \(\dfrac{a^2+2ab+b^2}{b}\) = \(\dfrac{\left(a+b\right)^2}{b}\)
c) \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\) = \(3+2-5=0\)
d) \(3+\sqrt{18}+\sqrt{3}+\sqrt{8}\) = \(3+3\sqrt{2}+\sqrt{3}+2\sqrt{2}\)
= \(3+\sqrt{3}+5\sqrt{2}\)
Tâm O của đường tròn nội tiếp tam giác đều cũng là giao điểm ba đường trung tuyến, ba đường cao.
Do đó đường cao h=AE=3.OE=3cm.
Trong tam giác đều, h = a√3/2 (a là độ dài mỗi cạnh).
Suy ra 
Do đó diện tích tam giác ABC là
Ta chọn (D).
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
a)\(\sqrt{\dfrac{4}{9-4\sqrt{5}}}-\sqrt{\dfrac{4}{9+4\sqrt{5}}} \Leftrightarrow \dfrac{\sqrt{4}}{\sqrt{(2-\sqrt{5}})^{2}}-\dfrac{\sqrt{4}}{(2+\sqrt{5})^{2}} \Leftrightarrow \dfrac{2(2+\sqrt{5})}{(\sqrt{5}-2)(2+\sqrt{5})}-\dfrac{2(\sqrt{5}-2)}{(\sqrt{5}-2)(2+\sqrt{5})} \Leftrightarrow \dfrac{4+2\sqrt{5}-(2\sqrt{5}-4)}{4-5} \Leftrightarrow \dfrac{8}{-1} = -8\)b)\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}} =\dfrac{\sqrt{2}\sqrt{8-4\sqrt{3}}}{\sqrt{2}\sqrt{2}} =\dfrac{\sqrt{16-8\sqrt{3}}}{2} =\dfrac{\sqrt{(2-2\sqrt{3})^{2}}}{2} =\dfrac{2\sqrt{3}-2}{2} =\dfrac{2(\sqrt{3}-1)}{2} =\sqrt{3}-1\)c)\(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}} =\sqrt{2}\sqrt{7-4\sqrt{3}}-\sqrt{2}\sqrt{12+6\sqrt{3}} =\sqrt{2}(\sqrt{(4-\sqrt{3})^{2}}-\sqrt{(3+\sqrt{3})^{2}}) =\sqrt{2}((4-\sqrt{3})-(3+\sqrt{3})) =\sqrt{2}(1-2\sqrt{3}) =\sqrt{2}-2\sqrt{6}\)
1.\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}=3\sqrt{2}\)
2.\(=5\sqrt{5}+4\sqrt{5}-9\sqrt{5}=0\)
a: \(=2\sqrt{2}+30\sqrt{2}-3\sqrt{2}+6\sqrt{2}=26\sqrt{2}\)
b: \(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}+\sqrt{3}+\dfrac{5}{2}\sqrt{3}=-\dfrac{9}{2}\sqrt{3}\)
a) ...= \(\dfrac{1}{4}\).\(6\sqrt{5}\) +\(2\sqrt{5}\) - \(3\sqrt{5}\) +5
= \(\dfrac{3}{2}\sqrt{5}\) -\(\sqrt{5}\) +5
=5 - \(\dfrac{1}{2}\sqrt{5}\)
d) ...= \(\sqrt{\dfrac{a}{\left(1+b\right)^2}}\) . \(\sqrt{\dfrac{4a\left(1+b\right)^2}{15^2}}\)
= \(\sqrt{\dfrac{4a^2\left(1+b\right)^2}{\left(1+b\right)^2.15^2}}\) = \(\sqrt{\dfrac{4a^2}{15^2}}\)= \(\dfrac{2a}{15}\)
Câu b nhé:
Ta có:
\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)
a) \(\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{1}{2}\)
b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=1+\sqrt{2}\)
d) \(\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\\ =\sqrt{81-17}=\sqrt{64}=8\)
\(a.\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(b.\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
\(c.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\dfrac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
\(d.\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}=\sqrt{81-17}=8\)
Chọn C