Question

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

Answer 1

#1.

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

$></p> <p>Найдем нули:</p> <p><img src=$

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

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

Answer 2

Let's solve these two equations step-by-step:

1. (x^2 - 3x - 1)^2 >= (x^2 + 7x + 1)^2

To solve this inequality, we'll first expand both sides and then try to simplify it:

Expanding the left side: (x^2 - 3x - 1)^2 = (x^2 - 3x - 1) * (x^2 - 3x - 1) = x^4 - 6x^3 + 11x^2 - 6x + 1

Expanding the right side: (x^2 + 7x + 1)^2 = (x^2 + 7x + 1) * (x^2 + 7x + 1) = x^4 + 14x^3 + 50x^2 + 14x + 1

Now the inequality becomes: x^4 - 6x^3 + 11x^2 - 6x + 1 >= x^4 + 14x^3 + 50x^2 + 14x + 1

Subtracting x^4 from both sides to simplify: -6x^3 + 11x^2 - 6x >= 14x^3 + 50x^2 + 14x

Combining like terms: -20x^3 - 39x^2 - 20x >= 0

Now, let's find the critical points by setting the left-hand side to zero: -20x^3 - 39x^2 - 20x = 0

To solve this equation, we can try to factor out common terms. One possible factorization is: x(-20x^2 - 39x - 20) = 0

Now, let's solve for x by setting each factor to zero:

1. x = 0
2. -20x^2 - 39x - 20 = 0

To solve the quadratic equation -20x^2 - 39x - 20 = 0, we can use factoring or the quadratic formula. Factoring it, we get: (-4x + 1)(5x + 20) = 0

Setting each factor to zero: 3) -4x + 1 = 0 x = 1/4

4. 5x + 20 = 0 x = -4

Now, we have four critical points: x = 0, x = 1/4, x = -4, and the solution to the inequality will lie in the intervals defined by these critical points. We can use test points to check which intervals satisfy the inequality.

2. { x^2 + 2xy = -1, x - 5y = 6 }

This appears to be a system of two equations. To solve it, we can use the method of substitution or elimination.

Let's start with substitution: From the second equation, we can express x in terms of y: x = 6 + 5y

Now, we'll substitute this value of x into the first equation: (6 + 5y)^2 + 2(6 + 5y)y = -1

Expanding and simplifying: 36 + 60y + 25y^2 + 12y + 10y^2 = -1 35y^2 + 72y + 37 = 0

Now we have a quadratic equation in terms of y. We can solve for y using the quadratic formula: y = (-b ± √(b^2 - 4ac)) / 2a

For this equation, a = 35, b = 72, and c = 37. y = (-72 ± √(72^2 - 4 * 35 * 37)) / 2 * 35

y = (-72 ± √(5184 - 5180)) / 70 y = (-72 ± √4) / 70 y = (-72 ± 2) / 70

Now, two possible values of y:

1. y = (-72 + 2) / 70 y = -70 / 70 y = -1

2. y = (-72 - 2) / 70 y = -74 / 70 y = -37 / 35

Now that we have the values of y, we can find the corresponding values of x using the second equation: For y = -1: x - 5(-1) = 6 x + 5 = 6 x = 6 - 5 x = 1

For y = -37 / 35: x - 5(-37 / 35) = 6 x + 37 / 7 = 6 x = 6 - 37 / 7 x = (42 - 37) / 7 x = 5 / 7

So, the solutions to the system are x = 1, y = -1, and x = 5/7, y = -37/35.