Toán Lớp 9: $\frac{4}{√3 -1}$ – $\frac{6}{√3}$ +$\frac{3+3 √3}{ √3+1}$
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Toán Lớp 9: $\frac{4}{√3 -1}$ – $\frac{6}{√3}$ +$\frac{3+3 √3}{ √3+1}$
Comments ( 2 )
=\frac{4(\sqrt{3} +1)}{(\sqrt{3})^2 -1}-\frac{2(\sqrt{3})^2}{\sqrt{3}}+ \frac{3(\sqrt{3}+1)}{\sqrt{3}+1}
=\frac{4(\sqrt{3} +1)}{3-1}-2\sqrt{3}+ 3
=\frac{4(\sqrt{3} +1)}{2}-2\sqrt{3}+ 3
=2(\sqrt{3} +1)-2\sqrt{3}+ 3
=2\sqrt{3} +2-2\sqrt{3}+ 3
= 5