В порядке убывания, значит, с большего до меньшего 1/2, 1/4, 1/5, 1/7, 1/9, 1/10, 1/13
![\sin^3x+3\cos^3x=2\cos x\\ \\ \sin^3x+\cos^3x+2\cos x(\cos^2x-1)=0\\ \\ \sin^2x(\sin x-\cos x)+\cos x(\cos^2x-\sin^2x)=0\\ \\ \sin^2x(\sin x-\cos x)-\cos x(\sin x-\cos x)(\sin x+\cos x)=0\\ \\ (\sin x-\cos x)(\sin^2 x-\cos x\sin x-\cos ^2x)=0](https://tex.z-dn.net/?f=%5Csin%5E3x%2B3%5Ccos%5E3x%3D2%5Ccos+x%5C%5C+%5C%5C+%5Csin%5E3x%2B%5Ccos%5E3x%2B2%5Ccos+x%28%5Ccos%5E2x-1%29%3D0%5C%5C+%5C%5C+%5Csin%5E2x%28%5Csin+x-%5Ccos+x%29%2B%5Ccos+x%28%5Ccos%5E2x-%5Csin%5E2x%29%3D0%5C%5C+%5C%5C+%5Csin%5E2x%28%5Csin+x-%5Ccos+x%29-%5Ccos+x%28%5Csin+x-%5Ccos+x%29%28%5Csin+x%2B%5Ccos+x%29%3D0%5C%5C+%5C%5C+%28%5Csin+x-%5Ccos+x%29%28%5Csin%5E2+x-%5Ccos+x%5Csin+x-%5Ccos+%5E2x%29%3D0)
Произведение равно нулю, если хотя бы один из множителей равен нулю
![\sin x-\cos x=0~~~|:\cos x\ne 0\\ \\ tgx-1=0\\ \\ tgx=1\\ \\ x= \frac{\pi}{4}+ \pi n,n \in \mathbb{Z}\\ \\ \sin^2x-\sin x\cos x-\cos^2x=0\\ \\ tg^2x-tgx-1=0](https://tex.z-dn.net/?f=%5Csin+x-%5Ccos+x%3D0~~~%7C%3A%5Ccos+x%5Cne+0%5C%5C+%5C%5C+tgx-1%3D0%5C%5C+%5C%5C+tgx%3D1%5C%5C+%5C%5C+x%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D%2B+%5Cpi+n%2Cn+%5Cin+%5Cmathbb%7BZ%7D%5C%5C+%5C%5C+%5Csin%5E2x-%5Csin+x%5Ccos+x-%5Ccos%5E2x%3D0%5C%5C+%5C%5C+tg%5E2x-tgx-1%3D0+)
Решим как квадратное уравнение относительно tg x.
![D=1+4=5\\ \\ tgx= \dfrac{1\pm \sqrt{5} }{2} ;~~~~\Rightarrow~~~~ x=arctg\bigg( \dfrac{1\pm \sqrt{5} }{2} \bigg)+ \pi n,n \in \mathbb{Z}](https://tex.z-dn.net/?f=D%3D1%2B4%3D5%5C%5C+%5C%5C+tgx%3D+%5Cdfrac%7B1%5Cpm+%5Csqrt%7B5%7D+%7D%7B2%7D+%3B~~~~%5CRightarrow~~~~+x%3Darctg%5Cbigg%28+%5Cdfrac%7B1%5Cpm+%5Csqrt%7B5%7D+%7D%7B2%7D+%5Cbigg%29%2B+%5Cpi+n%2Cn+%5Cin+%5Cmathbb%7BZ%7D)
100 км \ 50 = 2 часа бежал первый заяц
44 * 2 = 88 км пробежал второй заяц
А) 3 4/15 + у = 7 11/15
у = 7 11/15 - 3 4/15
у = 4 7/15
Проверка: 3 4/15 + 4 7/15 = 7 11/15
б) 5 4/13 - х = 4 5/13
х = 5 4/13 - 4 5/13
х = 4 17/13 - 4 5/13
х = 12/13
Проверка: 5 4/13 - 12/13 = 4 17/13 - 12/13 = 4 5/13