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+) \(\left(\sqrt{4}-\sqrt{3}\right)^2=4-2\sqrt{4\cdot3}+3=7-2\sqrt{7}=\sqrt{49}-\sqrt{48}\)
+) \(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)^2+6\sqrt{6}\)
\(=4\sqrt{2}-6\sqrt{6}+9-4\sqrt{2}+6\sqrt{6}\)
\(=9\)
+) Sửa : \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5-2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(=-2\sqrt{3}\)
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
Ta có
:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=|2-\sqrt{5}|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2=VP\left(đpcm\right)\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
Ta có:
\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}\)
\(=\frac{3+2\sqrt{2}}{2-1}\)
\(=3+2\sqrt{2}=VP\left(đpcm\right)\)
c,Bạn xem lại đề
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
Ta có:
\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)
\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)
\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)
\(=\frac{8}{5-4}\)
\(=8=VP\left(đpcm\right)\)
\(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
\(=9=VP\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
a) \(VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
=\(\sqrt{9^2-\left(\sqrt{17}\right)^2}=\sqrt{81-17}=\sqrt{64}=8=VP\)
b) \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
=\(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9=VP\)
VT = \(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)^2+6\sqrt{6}\)
\(=4\sqrt{2}-6\sqrt{6}+1-4\sqrt{2}+8+6\sqrt{6}=9\)=VP (đpcm)