[tex]\it (\sqrt{27}-4)(3\sqrt3+4)=(\sqrt{27}-4)(\sqrt{3^2\cdot3}+4)=(\sqrt{27}-4)(\sqrt{27}+4)=\\\\(\sqrt{27})^2-4^2=27-16=11[/tex]
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[tex]\it(\sqrt5-2)^2+4(\sqrt5-1)=(\sqrt5)^2-2\cdot\sqrt5\cdot2+2^2+4\sqrt5-4=\\\\=5-4\sqrt5+4+4\sqrt5-4=5[/tex]
