1/2(cos70-cos90)1/2(cos50-cos90)=1/4cos50cos70
cos90=0
cos70cos50=cos70cos50
А+6х/а : ах+6х²/а² =(<span>а+6х)/а * а</span>²<span> /х*(а+6х</span>) =а/х
-64/-64=1
Фотофотофотофотофотофотофото
![y(x)= \sqrt{2+sin^4(x)-cos(2x)} +\sqrt{2+cos^4(x)+cos(2x)} =\\ = \sqrt{2+(sin^2(x))^2-(1-2sin^2(x))} +\\+\sqrt{2+(cos^2(x))^2+(1-2sin^2(x))} =\\ = \sqrt{2+(sin^2(x))^2-1+2sin^2(x)} +\\+\sqrt{2+(1-sin^2(x))^2+1-2sin^2(x)} =\\ = \sqrt{1^2+2*1*sin^2(x)+(sin^2(x))^2} +\\+\sqrt{2+1-2sin^2(x)+sin^4(x)+1-2sin^2(x)} =\\ = \sqrt{(1+sin^2(x))^2} +\sqrt{2^2-2*2*sin^2(x)+(sin^2(x))^2} =\\ =\|1+sin^2(x)| +|2-sin^2(x)|=1+sin^2(x) +2-sin^2(x)=\\ =3](https://tex.z-dn.net/?f=y%28x%29%3D+%5Csqrt%7B2%2Bsin%5E4%28x%29-cos%282x%29%7D+%2B%5Csqrt%7B2%2Bcos%5E4%28x%29%2Bcos%282x%29%7D+%3D%5C%5C+%3D+%5Csqrt%7B2%2B%28sin%5E2%28x%29%29%5E2-%281-2sin%5E2%28x%29%29%7D+%2B%5C%5C%2B%5Csqrt%7B2%2B%28cos%5E2%28x%29%29%5E2%2B%281-2sin%5E2%28x%29%29%7D+%3D%5C%5C+%3D+%5Csqrt%7B2%2B%28sin%5E2%28x%29%29%5E2-1%2B2sin%5E2%28x%29%7D+%2B%5C%5C%2B%5Csqrt%7B2%2B%281-sin%5E2%28x%29%29%5E2%2B1-2sin%5E2%28x%29%7D+%3D%5C%5C+%3D+%5Csqrt%7B1%5E2%2B2%2A1%2Asin%5E2%28x%29%2B%28sin%5E2%28x%29%29%5E2%7D+%2B%5C%5C%2B%5Csqrt%7B2%2B1-2sin%5E2%28x%29%2Bsin%5E4%28x%29%2B1-2sin%5E2%28x%29%7D+%3D%5C%5C+%3D+%5Csqrt%7B%281%2Bsin%5E2%28x%29%29%5E2%7D+%2B%5Csqrt%7B2%5E2-2%2A2%2Asin%5E2%28x%29%2B%28sin%5E2%28x%29%29%5E2%7D+%3D%5C%5C+%3D%5C%7C1%2Bsin%5E2%28x%29%7C+%2B%7C2-sin%5E2%28x%29%7C%3D1%2Bsin%5E2%28x%29+%2B2-sin%5E2%28x%29%3D%5C%5C+%3D3)
Итого, графиком этой функции есть паралельная оси ОХ линия на высоте
![3](https://tex.z-dn.net/?f=3)
над ней:
Объяснение:
Представим уравнение в виде полного квадрата:
![x^2-6x-5 = x^2 - 6x + 9 -14 = (x - 3)^2 - 14](https://tex.z-dn.net/?f=x%5E2-6x-5%20%3D%20x%5E2%20-%206x%20%2B%209%20-14%20%3D%20%28x%20-%203%29%5E2%20-%2014)
То есть это парабола y=x², смещенная на 3 единицы вправо и на 14 единиц вниз