Lucrez dintr-o culegere pentru admitere la facultatea de politehnica din Timisoara, si nu reusesc sa ii dau de cap acestui exercitiu. Dau 50+25 puncte.
Nustiucesapunaici
Rezolvarea este in fotografia de mai jos.
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matepentrutoti
![\displaystyle \frac{sin^6x+cos^6x-1}{sin^4x+cos^4x-1} =\\ = \frac{(sin^2x)^3+(cos^2x)^3-1}{(sin^2x+cos^2x)^2-2sin^2xcos^2x-1}=\\ = \frac{(sin^2x+cos^2x)[(sin^2x)^2-sin^2xcos^2x+(cos^2x)^2]-1}{-2sin^2xcos^2x}=\\ = \frac{(sin^2x+cos^2x)^2-3sin^2xcos^2x-1}{-2sin^2xcos^2x} =\\ = \frac{-3sin^2xcos^2x}{-2sin^2xcos^2x} = \frac{3}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Cfrac%7Bsin%5E6x%2Bcos%5E6x-1%7D%7Bsin%5E4x%2Bcos%5E4x-1%7D+%3D%5C%5C%0A%3D+%5Cfrac%7B%28sin%5E2x%29%5E3%2B%28cos%5E2x%29%5E3-1%7D%7B%28sin%5E2x%2Bcos%5E2x%29%5E2-2sin%5E2xcos%5E2x-1%7D%3D%5C%5C%0A%3D+%5Cfrac%7B%28sin%5E2x%2Bcos%5E2x%29%5B%28sin%5E2x%29%5E2-sin%5E2xcos%5E2x%2B%28cos%5E2x%29%5E2%5D-1%7D%7B-2sin%5E2xcos%5E2x%7D%3D%5C%5C%0A%3D+++%5Cfrac%7B%28sin%5E2x%2Bcos%5E2x%29%5E2-3sin%5E2xcos%5E2x-1%7D%7B-2sin%5E2xcos%5E2x%7D+%3D%5C%5C%0A%3D+%5Cfrac%7B-3sin%5E2xcos%5E2x%7D%7B-2sin%5E2xcos%5E2x%7D+%3D+%5Cfrac%7B3%7D%7B2%7D+ "\displaystyle \frac{sin^6x+cos^6x-1}{sin^4x+cos^4x-1} =\\ = \frac{(sin^2x)^3+(cos^2x)^3-1}{(sin^2x+cos^2x)^2-2sin^2xcos^2x-1}=\\ = \frac{(sin^2x+cos^2x)[(sin^2x)^2-sin^2xcos^2x+(cos^2x)^2]-1}{-2sin^2xcos^2x}=\\ = \frac{(sin^2x+cos^2x)^2-3sin^2xcos^2x-1}{-2sin^2xcos^2x} =\\ = \frac{-3sin^2xcos^2x}{-2sin^2xcos^2x} = \frac{3}{2}")
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