(2x-3y)y+(2y-3x)x = 2xy-3y^2+2xy - 3x^2 = 4xy - 3*(x^2+y^2) = 4*(-5) -3*(x^2+y^2) = -20-3*131 = -413<span>
xy = -5
x+y = -11
=> (x+y)^2 = 121
x^2 + 2xy+y^2 = 121
x^2+y^2 = 121-2xy = 121-2*(-5) = 121+10 = 131</span>
2y(n-m)+(m-n)=2y(n-m)-(n-m)=(n-m)(2y-1)
Решение
3log₇<span>X -l ogx7 = 1
ОДЗ: x > 0; x </span>∈ (0; + ∞)
3log₇<span> (x) - log</span>₇<span>7 / log</span>₇ (x)<span> = 1
</span>3log²₇ (x) + log₇ (x) - 1 = 0
log₇ (x) = t
3t² + t - 1 = 0
D = 1 + 4*3*1 = 13
t₁ = (- 1 - √13)/6 не удовлетворяет ОДЗ.
t₂ = (- 1 + √13)/6
log₇ (x) = (- 1 + √13)/6
x = [7^(- 1 + √13)/6]