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31 tháng 7 2019
\(\sqrt{13-2\sqrt{42}}=\sqrt{6-2\sqrt{6}.\sqrt{7}+7}=\sqrt{\left(\sqrt{6}-\sqrt{7}\right)^2}=\left|\sqrt{6}-\sqrt{7}\right|=\sqrt{7}-\sqrt{6}\)
\(\sqrt{46+6\sqrt{5}}=\sqrt{45+6\sqrt{5}+1}=\sqrt{3^2.5+6\sqrt{5}+1}=\sqrt{3^2.5+2.3.\sqrt{5}+1^2}=\sqrt{\left(3.\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)
\(\sqrt{12-3\sqrt{15}}=\sqrt{3}\sqrt{4-\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{8-2\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{3-2\sqrt{15}+5}=\sqrt{\frac{3}{2}}.\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\frac{3}{2}}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{15}+5}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{3}\sqrt{5}+5}-\sqrt{3+2\sqrt{3}\sqrt{5}+5}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}=-2\sqrt{3}\)
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)=\sqrt{\frac{1}{2}}\left(1+\sqrt{5}-\sqrt{5}+1\right)=\sqrt{\frac{1}{2}}.2=\sqrt{\frac{4}{2}}=\sqrt{2}\)
5 tháng 9 2020
a) Ta có: \(A=2\sqrt{2+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12}+1}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{3+1-2\sqrt{3}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{3}-1}\)
\(\Leftrightarrow A=2\sqrt{\sqrt{3}+1}\)
\(\Leftrightarrow A\approx3,30578\)
b) Ta có: \(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(\Leftrightarrow B=\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)
\(\Leftrightarrow B=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(\Leftrightarrow B=\sqrt{2}.\left(4-2\right)\)
\(\Leftrightarrow B=2\sqrt{2}\)
\(\Leftrightarrow B\approx2,82843\)
1 tháng 8 2018
\(a.\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\) \(b.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\) \(c.\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
8 tháng 6 2017
2) Dễ thấy\(\left(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)=x^2-6x+13-x^2+6x-10=3\)
\(\Leftrightarrow1.\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)=3\)
\(\Leftrightarrow\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}=3\)
9 tháng 6 2017
Ta có: a+ b= \(\frac{-1+\sqrt{2}}{2}\) + \(\frac{-1-\sqrt{2}}{2}\)= -1
a*b = \(\frac{-1+\sqrt{2}}{2}\)* \(\frac{-1-\sqrt{2}}{2}\)= -\(\frac{1}{4}\)
a2 + b2 = (a+ b)2 - 2ab = 1+ \(\frac{1}{2}\)= \(\frac{3}{2}\)
a4 + b4 = (a2 + b2 )2 - 2a2b2 = \(\frac{9}{4}\)- \(\frac{1}{8}\)= \(\frac{17}{8}\)
a3 + b3 = ( a + b)3 - 3ab(a + b ) = -1-\(\frac{3}{4}\)= \(\frac{-7}{4}\)
vay a7 + b7 = (a3 + b3 )(a4 + b4 ) -a3b3(a+b)= \(\frac{-7}{4}\)* \(\frac{17}{8}\)- (-\(\frac{1}{64}\)) * (-1) = \(\frac{-239}{64}\)
18 tháng 6 2017
a) \(=3-\sqrt{5}\)
b) \(=\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{12}-\sqrt{1}\right)^2}\)
\(=\left|\sqrt{3}-1\right|-\left|2\sqrt{3}-1\right|\)
\(=\sqrt{3}-1-\left(2\sqrt{3}-1\right)\)
\(=\sqrt{3}-1-2\sqrt{3}+1\)
\(=-\sqrt{3}\)
19 tháng 6 2017
đúng rồi yey, nhưng tớ trả biết phần a có dúng đề k nữa.
làm sao cô cho câu dễ thế dk
1 tháng 9 2018
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
1 tháng 9 2018
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
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