Решение
y=(tg3x)/(tg3x-1)
y` = [(3/cos²3x)*(tg3x - 1) - (3/cos²3x)*tg3x] / (tg3x-1)² =
= [(3/cos²3x) * (tg3x - 1 - tg3x)] / (tg3x-1)² =
= - (3/cos²3x) / (tg3x-1)²
1.
а = с * sinα,
a = 26 * sin 67°23' = 26 * 0,9231 = 24,
b = c * cosα,
b = 26 * 0,3846 = 10,
2.
c = b : sinβ,
c = 28 : 0,5282 = 53,
a = √(c² - b²),
a = √(53² - 28²) = √(2809 - 784) = √2025 = 45,
3.
b = √(c² - a²),
b = √(29² - 21²) = √(841 - 441) = √400 = 20,
4.
c = b : cosα,
c = 63 : 0,9689 = 65,
a = √(c² - b²),
a = √(65² - 63²) = √(4225 - 3969) = √256 = 16
X+3y=-1
x²+2xy+y=3
x=-1-3y(-1-3y)^2+2y(-1-3y)+y=31+6y+9y^2-2y-6y^2+y-3=0
сокращаем
3y^2+5y-2=0
дискриминант=25+24=49=7^2
y1=(-5+7)/6=1/3
y2=(-5-7)/6=-2
так как x=-1-3y , то подставляемx
1=-1-3*1/3=-2 <span>x2=-1-3(-2)=5
Вот )
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