Question

Найдите значения выражений:

Answer 1

$1) \ \sqrt{(2Cos30^\circ+1)^{2} } -\sqrt{(1-2Sin60^\circ)^{2}}=|2Cos30^\circ+1|-|1-2Sin60^\circ|=\\\\=\Big|2\cdot\dfrac{\sqrt{3} }{2}+1\Big|-\Big|1-2\cdot \dfrac{\sqrt{3} }{2}\Big|=\sqrt{3} +1-(\sqrt{3}-1)=\sqrt{3} +1-\sqrt{3}+1=\boxed2$

$2) \ \dfrac{2tg\frac{\pi }{4}-Sin\frac{3\pi }{2}}{(tg\frac{\pi }{6}-tg0)Cos\frac{\pi }{6} }=\dfrac{2\cdot 1-(-1)}{(\frac{1}{\sqrt{3} }-0)\cdot\frac{\sqrt{3} }{2} }=\dfrac{2+1}{\frac{1}{2} } =3\cdot 2=\boxed6$

Answer 2

Ответ:

$1)\ \ \sqrt{(2\, cos30^\circ +1)^2}-\sqrt{(1-2sin60^\circ )^2}=|\ 2cos30^\circ +1\ |-|\ 1-2sin60^\circ \ |=\\\\=\Big|\ 2\cdot \dfrac{\sqrt3}{2}+1\ \Big|-\Big|\ 1-2\cdot \dfrac{\sqrt3}{2}\ \Big|=|\sqrt3+1|-|1-\sqrt3|=(\sqrt3+1)-(\sqrt3-1)=\\\\=\sqrt3+1-\sqrt3+1=2$

$2)\ \ \dfrac{2\, tg\dfrac{\pi}{4}-sin\dfrac{3\pi}{2}}{\Big(tg\dfrac{\pi}{6}-tg0\Big)\cdot cos\dfrac{\pi}{6}}=\dfrac{2\cdot 1-(-1)}{\Big(\dfrac{\sqrt3}{3}-0\Big)\cdot \dfrac{\sqrt3}{2}}=\dfrac{3}{\dfrac{3}{3\cdot 2}}=3\cdot 2=6$