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Lời giải:
$\sqrt{12}-\sqrt{27}+\sqrt{75}=\sqrt{2^2.3}-\sqrt{3^2.3}+\sqrt{5^2.3}$
$=2\sqrt{3}-3\sqrt{3}+5\sqrt{3}=(2-3+5)\sqrt{3}=4\sqrt{3}$
Đáp án D
a/\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
b/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)
c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)
\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)
\(=-4\sqrt{5}\)
√163 ≈ 12.767 √127 ≈ 11.269 √475 ≈ 21.794
2√163 - 3√127 - 6√475 ≈ 2(12.767) - 3(11.269) - 6(21.794)
≈ 25.534 - 33.807 - 130.764
≈ -138.037
Vậy giá trị của biểu thức 2√163 - 3√127 - 6√475 là -138.037.
\(=2\cdot\dfrac{4}{\sqrt{3}}-3\cdot\dfrac{1}{3\sqrt{3}}-6\cdot\dfrac{2}{5\sqrt{3}}\)
\(=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}=\dfrac{7}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}\left(7-\dfrac{12}{5}\right)=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
\(=2\cdot\dfrac{4}{\sqrt{3}}-3\cdot\dfrac{1}{3\sqrt{3}}-6\cdot\dfrac{2}{3\sqrt{5}}\\ =\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{4}{\sqrt{5}}=\dfrac{7}{\sqrt{3}}-\dfrac{4\sqrt{5}}{5}\\ =\dfrac{7\sqrt{3}}{3}-\dfrac{4\sqrt{5}}{5}=\dfrac{35\sqrt{3}-12\sqrt{5}}{15}\)
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
3v3(1-v3) / 15v3= (1-v3)/5