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\(\left(5+\frac{2\sqrt{6}}{\sqrt{3}}+\sqrt{2}\right)-\left(5-\frac{2\sqrt{6}}{\sqrt{3}}-\sqrt{2}\right)\)
=\(5+\frac{2\sqrt{6}}{\sqrt{3}}+\sqrt{2}-5+\frac{2\sqrt{6}}{\sqrt{3}}+\sqrt{2}\)
=\(\left(5-5\right)+\left(\frac{2\sqrt{6}}{\sqrt{3}}+\frac{2\sqrt{6}}{\sqrt{3}}\right)+\left(\sqrt{2}+\sqrt{2}\right)\)
=\(0+\frac{4\sqrt{6}}{\sqrt{3}}+2\sqrt{2}\)
=\(\frac{4\sqrt{2}.\sqrt{3}}{\sqrt{3}}+2\sqrt{2}\)
=\(4\sqrt{2}+2\sqrt{2}\)
=\(6\sqrt{2}\)
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
a)
\(2\sqrt{5}\)+ I1-\(\sqrt{5}\)I
\(2\sqrt{5}\)+1-\(\sqrt{5}\)
1+\(\sqrt{5}\)
b: \(=\dfrac{\sqrt{3}-1+\sqrt{3}+1-4\sqrt{3}}{2}=-\sqrt{3}\)
vãi bạn viết dấu căn thế ai hiểu đc