Question

Срочноо Решитеее!!!
Даю 20 баллов!!!
$(2 - \sqrt{3} ) {}^{2} (7 + 4 \sqrt{3} ) - 4 \sqrt{3 \times \frac{1}{16} }$
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Answer 1

$(2-\sqrt3)^2\, (7+4\sqrt3)-4\sqrt{3\cdot \frac{1}{16}}=(4-4\sqrt3+3)\, (7+4\sqrt3)-4\cdot \frac{1}{4}\sqrt3=\\\\=(7-4\sqrt3)(7+4\sqrt3)-\sqrt3=7^2-(4\sqrt3)^2-\sqrt3=49-16\cdot 3-\sqrt3=\\\\=49-48-\sqrt3=1-\sqrt3$

Answer 2

$(2-\sqrt{3})^{2}(7+4\sqrt{3})-4*\sqrt{3*\frac{1}{16}}=(2-\sqrt{3})^{2}*(4+4\sqrt{3}+3)-4*\sqrt{\frac{3}{16}}=(2-\sqrt{3})^{2}*(2^{2}+2*2*\sqrt{3}+(\sqrt{3})^{2})-4*\frac{\sqrt{3} }{4}=(2-\sqrt{3})^{2}*(2+\sqrt{3})^{2}-\sqrt{3}=[2^{2}-(\sqrt{3})^{2}]^{2} -\sqrt{3}=(4-3)^{2}-\sqrt{3}=1-\sqrt{3}\\\\Otvet:\boxed{1-\sqrt{3}}$