Question

1. (x^2 - 3x - 1)^2 >= ( x^2 + 7x + 1 )^2
2. { x^2 + 2xy = -1
x - 5y = 6

Answer 1

#1.

$(x^2 - 3x - 1)^2 \geq (x^2 +7x+1)^2$

$(x^2 - 3x - 1)^2 - (x^2 +7x+1)^2\geq0\\((x^2 - 3x - 1) - (x^2 +7x+1))((x^2 - 3x - 1) + (x^2 +7x+1)) \geq 0\\(x^2 - 3x - 1 -x^2 - 7x - 1)(x^2 - 3x - 1 + x^2 + 7x + 1) \geq 0\\(-10x - 2)(2x^2 +4x)\geq 0$

Найдем нули:

$\left[\begin{array}{ccc}-10x - 2=0\\2x^2 +4x = 0\\\end{array}\right$

$\left[\begin{array}{ccc}x =-0,2\\x = 0\\x = -2\end{array}\right$

x ∈ (-∞; -2] ∪ [-0,2; 0]

Ответ: x ∈ (-∞; -2] ∪ [-0,2; 0].

#2.

$\left \{ {{x^2 + 2xy=-1} \atop {x-5y=6}} \right.$

$\left \{ {{x=6 + 5y} \atop {(6+5y)^2 + 2y(6+5y)=-1}} \right.$

$(6+5y)(6+5y + 2y) = -1\\(6+5y)(6+7y) = -1\\36 + 42y + 30y + 35y^2 + 1 = 0\\35y^2 + 72y + 37 = 0 \\35y^2 + 37y + 35y + 37 = 0\\y (35y + 37) + 35y + 37 = 0\\(35y + 37)(y + 1) = 0\\\left \{ {{y_1=-1} \atop {y_2 = \frac{-35}{37} }} \right.$

$\left[\begin{array}{cccc}y_1 = -1\\x_1 = 1\\y_2 = \frac{-37}{35} \\x_2 = \frac{5}{7} \end{array}\right$

Ответ: $(x_1;y_1) = (\frac{5}{7} ; -\frac{37}{35} ); (x_2;y_2) = (1; -1)$