Question

(2x+3)(3x+1)+(5x+2)(2x+5)=4x-1

Answer 1

(2x+3)(3x+1)+(5x+2)(2x+5)=4x-1

6х²+2х+9х+3+10х²+10х+4х+10=4х-1

16х²+12х+13-1+4х-4х=0

16х²+12х+12=0

квадратное уравнение, решаем через Д

Д = 12²-4*16*12 = -624

Т.к. Д меньше 0, то уравнение не имеет вещественных корней, но комплексные корни имеются:

х₁=$-\frac{3}{8} +\frac{\sqrt[]{39i} }{8}$

х₂=$-\frac{3}{8} -\frac{\sqrt[]{39i} }{8}$

Answer 2

(2х+3)(3х+1)+(5х+2)(2х+5)=4х-1 6х^2+2х+9х+3+10х^2+25х+4х+10-4х+1=0 16х^2+36х+14=0, делим на 2 8х^2+18х+7=0 Д=324-224=100 Х=(-18+100)/16=5 1/8 Х=(-18-100)/16=7 3/20

Answer 3

Let's verify if the given equation is true.

First, let's expand the left-hand side (LHS) of the equation:

(LHS) = (2x+3)(3x+1) + (5x+2)(2x+5)

To expand each bracket, we use the distributive property:

(LHS) = 2x * 3x + 2x * 1 + 3 * 3x + 3 * 1 + 5x * 2x + 5x * 5 + 2 * 2x + 2 * 5

Now, simplify each term:

(LHS) = 6x^2 + 2x + 9x + 3 + 10x^2 + 25 + 4x + 10

Combine like terms:

(LHS) = 6x^2 + 10x^2 + 2x + 9x + 4x + 3 + 25 + 10

(LHS) = 16x^2 + 15x + 38

Now, the equation becomes:

16x^2 + 15x + 38 = 4x - 1

To solve for x, move all the terms to one side of the equation:

16x^2 + 15x + 38 - (4x - 1) = 0

Combine like terms:

16x^2 + 15x + 38 - 4x + 1 = 0

Now, simplify:

16x^2 + 11x + 39 = 0

Unfortunately, the equation we ended up with is not equal to the original equation (4x - 1). Therefore, the given equation is not true for all values of x.