Это уравнение неразрешимо в рациональных числах.
Но если не обращать, на это внимания и взять иррациональные стороны, то это остроугольный треугольник.
Потому что, если a^3 + b^3 = c^3, то a^2 + b^2 < c^2
![f(x) = 9^x + 5*3^{-2x} = 3^{2x} + 5*3^{-2x}\\\\ f'(x) = 2*3^{2x}ln3 -10*3^{-2x}ln3\\\\](https://tex.z-dn.net/?f=f%28x%29+%3D+9%5Ex+%2B+5%2A3%5E%7B-2x%7D+%3D+3%5E%7B2x%7D+%2B+5%2A3%5E%7B-2x%7D%5C%5C%5C%5C+f%27%28x%29+%3D+2%2A3%5E%7B2x%7Dln3+-10%2A3%5E%7B-2x%7Dln3%5C%5C%5C%5C)
![f'(x) = 0; \ 2*3^{2x}ln3 -10*3^{-2x}ln3 = 0\\\\3^{2x} -5*3^{-2x} = 0\\\\ 3^{2x} = 5*3^{-2x}\\\\ 1/5 = 3^{-4x}, \ 3^{4x} = 5, \ 81^x = 5, \ x = log_{81}5\\](https://tex.z-dn.net/?f=f%27%28x%29+%3D+0%3B+%5C+2%2A3%5E%7B2x%7Dln3+-10%2A3%5E%7B-2x%7Dln3+%3D+0%5C%5C%5C%5C3%5E%7B2x%7D+-5%2A3%5E%7B-2x%7D+%3D+0%5C%5C%5C%5C+3%5E%7B2x%7D+%3D+5%2A3%5E%7B-2x%7D%5C%5C%5C%5C+1%2F5+%3D+3%5E%7B-4x%7D%2C+%5C+3%5E%7B4x%7D+%3D+5%2C+%5C+81%5Ex+%3D+5%2C+%5C+x+%3D+log_%7B81%7D5%5C%5C)
![f(x) < 0, x < log_{81}5,\\\\ f(x) > 0, x > log_{81}5\\\\](https://tex.z-dn.net/?f=f%28x%29+%3C+0%2C+x+%3C+log_%7B81%7D5%2C%5C%5C%5C%5C+f%28x%29+%3E+0%2C+x+%3E+log_%7B81%7D5%5C%5C%5C%5C)
Функция f(x) имеет в точке x = log_{81}5, минимум.
![f(log_{81}5) = 3^{2log_{81}5} + 5*3^{-2log_{81}5} = 3^{2log_{3^4}5} + 5*3^{-2log_{3^4}5} =\\\\ 3^{log_{3}\sqrt{5}} + 5*3^{log_{3}(1/\sqrt{5})} = \sqrt{5} + 5/\sqrt{5} = 2\sqrt{5}](https://tex.z-dn.net/?f=f%28log_%7B81%7D5%29+%3D+3%5E%7B2log_%7B81%7D5%7D+%2B+5%2A3%5E%7B-2log_%7B81%7D5%7D+%3D+3%5E%7B2log_%7B3%5E4%7D5%7D+%2B+5%2A3%5E%7B-2log_%7B3%5E4%7D5%7D+%3D%5C%5C%5C%5C+3%5E%7Blog_%7B3%7D%5Csqrt%7B5%7D%7D+%2B+5%2A3%5E%7Blog_%7B3%7D%281%2F%5Csqrt%7B5%7D%29%7D+%3D+%5Csqrt%7B5%7D+%2B+5%2F%5Csqrt%7B5%7D+%3D+2%5Csqrt%7B5%7D)
При x -> +∞, f(x) - > +∞, при x -> -∞, f(x) - > +∞
Mножество значений функции f(x): ![[2\sqrt{5}, +\infty)](https://tex.z-dn.net/?f=%5B2%5Csqrt%7B5%7D%2C+%2B%5Cinfty%29)
1) cos a cos 3a - sin a sin 3a = cos (a+3a) = cos 4a
2) sin 2a cos a + cos 2a sin a = sin (2a+a) = sin 3a
3) sin40°cos5°+cos40°sin5° = sin(40°+5°)= sin 45° = ![\frac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=+%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D+++)
4) ![\dfrac{tg\frac{7\pi}{15}-tg\frac{2\pi}{15}}{1+tg\frac{7\pi}{15}tg\frac{2\pi}{15}} = tg(\frac{7\pi}{15}-\frac{2\pi}{15}) =tg\frac{5\pi}{15}=tg\frac{\pi}{3}=\sqrt{3}](https://tex.z-dn.net/?f=+%5Cdfrac%7Btg%5Cfrac%7B7%5Cpi%7D%7B15%7D-tg%5Cfrac%7B2%5Cpi%7D%7B15%7D%7D%7B1%2Btg%5Cfrac%7B7%5Cpi%7D%7B15%7Dtg%5Cfrac%7B2%5Cpi%7D%7B15%7D%7D++%3D+tg%28%5Cfrac%7B7%5Cpi%7D%7B15%7D-%5Cfrac%7B2%5Cpi%7D%7B15%7D%29+%3Dtg%5Cfrac%7B5%5Cpi%7D%7B15%7D%3Dtg%5Cfrac%7B%5Cpi%7D%7B3%7D%3D%5Csqrt%7B3%7D++)
5) ![\pi <\alpha <\frac{3\pi}{2} => cos \alpha <0](https://tex.z-dn.net/?f=+%5Cpi+%3C%5Calpha+%3C%5Cfrac%7B3%5Cpi%7D%7B2%7D++%3D%3E+cos+%5Calpha+%3C0+)
![sin^2\alpha +cos^2\alpha =1\\ cos\alpha =-\sqrt{1-(\frac{4}{5})^2} =-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}](https://tex.z-dn.net/?f=+sin%5E2%5Calpha+%2Bcos%5E2%5Calpha+%3D1%5C%5C+cos%5Calpha+%3D-%5Csqrt%7B1-%28%5Cfrac%7B4%7D%7B5%7D%29%5E2%7D++%3D-%5Csqrt%7B1-%5Cfrac%7B16%7D%7B25%7D%7D%3D-%5Csqrt%7B%5Cfrac%7B9%7D%7B25%7D%7D%3D-%5Cfrac%7B3%7D%7B5%7D++++++)
Пусть 1 кг. абрикосов - х тенге,
тогда 1 кг. персиков - х+50 тенге,
4 кг. персиков - 4*(х+50) тенге,
6 кг. абрикосов - 6х тенге.
Составляем уравнение:
4*(х+50)+6х=4200
4х+200+6х=4200
10х=4200-200
10х=4000; х=4000:10; х=400 (тенге) - за 1 кг. абрикосов
400+50=450 (тенге) - за 1 кг. персиков
6*400=2400 (тенге) - за 6 кг. абрикосов
4*450=1800 (тенге) - за 4 кг. персиков
2Sin(7π/2 - х) Sinx = Cosx (7π/2 ; 5π)
-2CosxSinx -Cosx = 0
Cosx(2Sinx +1) = 0
Cosx = 0 или 2Sinx + 1 = 0
x = π/2 + πk , k ∈ Z Sinx = -1/2
х = (-1)ⁿ⁺¹π/6 + nπ, n ∈ Z
В указанный промежуток попадают числа : π/2 и 20π/6