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Ответ : Д .
![\frac{\sqrt2cos \alpha -2cos(45- \alpha )}{2sin(30+ \alpha )-\sqrt3sin \alpha }=[\, \frac{:2}{:2}\, ]=\frac{\frac{\sqrt2}{2}cos \alpha -cos(45- \alpha )}{sin(30+ \alpha )-\frac{\sqrt3}{2}sin \alpha }=\\\\=\frac{\frac{\sqrt2}{2}cos \alpha -(cos45cos \alpha +sin45sin \alpha )}{sin30cos \alpha +cos30sin \alpha -\frac{\sqrt3}{2}sin \alpha }=\frac{\frac{\sqrt2}{2}cos \alpha -\frac{\sqrt2}{2}cos \alpha -\frac{\sqrt2}{2}sin \alpha }{\frac{1}{2}cos \alpha +\frac{\sqrt3}{2}sin \alpha -\frac{\sqrt3}{2}sin \alpha } \frac{\sqrt2cos \alpha -2cos(45- \alpha )}{2sin(30+ \alpha )-\sqrt3sin \alpha }=[\, \frac{:2}{:2}\, ]=\frac{\frac{\sqrt2}{2}cos \alpha -cos(45- \alpha )}{sin(30+ \alpha )-\frac{\sqrt3}{2}sin \alpha }=\\\\=\frac{\frac{\sqrt2}{2}cos \alpha -(cos45cos \alpha +sin45sin \alpha )}{sin30cos \alpha +cos30sin \alpha -\frac{\sqrt3}{2}sin \alpha }=\frac{\frac{\sqrt2}{2}cos \alpha -\frac{\sqrt2}{2}cos \alpha -\frac{\sqrt2}{2}sin \alpha }{\frac{1}{2}cos \alpha +\frac{\sqrt3}{2}sin \alpha -\frac{\sqrt3}{2}sin \alpha }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt2cos+%5Calpha+-2cos%2845-+%5Calpha+%29%7D%7B2sin%2830%2B+%5Calpha+%29-%5Csqrt3sin+%5Calpha+%7D%3D%5B%5C%2C+%5Cfrac%7B%3A2%7D%7B%3A2%7D%5C%2C+%5D%3D%5Cfrac%7B%5Cfrac%7B%5Csqrt2%7D%7B2%7Dcos+%5Calpha+-cos%2845-+%5Calpha+%29%7D%7Bsin%2830%2B+%5Calpha+%29-%5Cfrac%7B%5Csqrt3%7D%7B2%7Dsin+%5Calpha+%7D%3D%5C%5C%5C%5C%3D%5Cfrac%7B%5Cfrac%7B%5Csqrt2%7D%7B2%7Dcos+%5Calpha+-%28cos45cos+%5Calpha+%2Bsin45sin+%5Calpha+%29%7D%7Bsin30cos+%5Calpha+%2Bcos30sin+%5Calpha+-%5Cfrac%7B%5Csqrt3%7D%7B2%7Dsin+%5Calpha+%7D%3D%5Cfrac%7B%5Cfrac%7B%5Csqrt2%7D%7B2%7Dcos+%5Calpha+-%5Cfrac%7B%5Csqrt2%7D%7B2%7Dcos+%5Calpha+-%5Cfrac%7B%5Csqrt2%7D%7B2%7Dsin+%5Calpha+%7D%7B%5Cfrac%7B1%7D%7B2%7Dcos+%5Calpha+%2B%5Cfrac%7B%5Csqrt3%7D%7B2%7Dsin+%5Calpha+-%5Cfrac%7B%5Csqrt3%7D%7B2%7Dsin+%5Calpha+%7D)
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