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MV
3 tháng 7 2018
a.
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\\ =\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\\ =\sqrt{81-17}\\ =\sqrt{64}\\=8\)
3 tháng 7 2018
\(a.VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=8=VP\)
\(b.\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=3\sqrt{3}-\sqrt{2}\) ( thiếu đề )
\(VT=\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=\dfrac{1}{3-2\sqrt{3}.\sqrt{2}+2}+\dfrac{2}{3+2\sqrt{3}.\sqrt{2}+2}=\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-2\sqrt{2}=3\sqrt{3}-\sqrt{2}=VP\)
12 tháng 7 2018
\(1.\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+2.2\sqrt{5}+4}}=\sqrt{17-4\left(\sqrt{5}+2\right)}=\sqrt{5-2.2\sqrt{5}+4}=\sqrt{5}-2\)
\(2.\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{17-6\sqrt{2+2\sqrt{2}+1}}=\sqrt{17-6\left(\sqrt{2}+1\right)}=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)\(3.\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}=\sqrt{3+\sqrt{3-2\sqrt{3}+1}}=\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(4.\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}=\sqrt{25+2.5\sqrt{2}+2}.\left(5-\sqrt{2}\right)=\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)=5-2=3\)
TH
16 tháng 7 2018
a) \(\sqrt{\left(\sqrt{3}-3\right)^2}-\sqrt{16+6\sqrt{3}}=3-\sqrt{3}-\sqrt{\left(3+\sqrt{3}\right)^2+4}\)
b) \(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{5}-5}{\sqrt{5}-1}=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\dfrac{2\left(2-\sqrt{2}\right)}{4-2}-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5-1}}=\sqrt{5}+\sqrt{2}+2-\sqrt{2}-\sqrt{5}=2\)
c) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}=2+\sqrt{17-4\left(\sqrt{5}+2\right)}=2+\sqrt{9-4\sqrt{5}}=2+\sqrt{5}-2=\sqrt{5}\)
d) \(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\cdot\dfrac{1}{\sqrt{3}}=\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\cdot\dfrac{1}{\sqrt{3}}=1\)
9 tháng 8 2018
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
\(\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}=\dfrac{2-\sqrt{5}}{\sqrt{9-4\sqrt{5}}}=\dfrac{2-\sqrt{5}}{\sqrt{5}-2}=-1\)
\(\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{9+4\sqrt{5}}}}=\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\\ =\dfrac{2-\sqrt{5}}{\sqrt{17-4\left(\sqrt{5}+2\right)}}=\dfrac{2-\sqrt{5}}{\sqrt{9-4\sqrt{5}}}\\ =\dfrac{2-\sqrt{5}}{\sqrt{\left(\sqrt{5}-2\right)^2}}=\dfrac{2-\sqrt{5}}{\sqrt{5}-2}=-1\)