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a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
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:
\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)
\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)
\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)
\(=\frac{8}{5-4}\)
\(=8=VP\left(đpcm\right)\)
a. Sửa đề: \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
biến đổi vế trái :
ta có :\(\left(3+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3^2-\left(\sqrt{5}\right)^2}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=2(\(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\))
=2(\(\sqrt{5}+5-\sqrt{5}-1\))
=2.4=8=VP
=> đpcm
b. Đặt vế trái là A
ta có \(A^2=\sqrt{2}+1-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\sqrt{2}-1\)
=\(2\sqrt{2}-2\)
=2\(\left(\sqrt{2}-1\right)\)
=> A=\(\sqrt{2\left(\sqrt{2}-1\right)}\)
vậy VT=VP =>đpcm
\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\frac{1}{10}}+10\right)=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\frac{3\sqrt{10}}{10}-10\)
\(=-3\sqrt{10}+10-\frac{3\sqrt{10}}{10}-10=-3\sqrt{10}-\frac{3\sqrt{10}}{10}=-3\sqrt{10}\left(1+\frac{1}{10}\right)=\frac{-33\sqrt{10}}{10}=-3,3\sqrt{10}\)