Question

18÷(х+1.5)=1.5÷(х-1.5)+30

Answer 1

18/(x+1,5)=1,5/(x-1,5)+30| умножим на (x-1,5)*(x+1,5) не равно 0
х^2-1,5^2 не равно 0
х^2 не равно 2,25
х1 не равно 2,25
х2 не равно -2,25
18(х-1,5)=1,5(х+1,5)+30(х^2-2,25)
18х-27=1,5х+2,25+30х^2-67,5
30х^2-18х+1,5х+2,25+27-67,5=0
30х^2-18х-38,25=0
D=(-18)^2-4*30*(-38,25)=324+4590=4914
x1=(-(-18)-V4914)/2*30=(18-70,1)/60=-52,1/60=-0,87
x2=(-(-18)+V4914)/2*30=(18+70,1)/60=88,1/60=1,47

Answer 2

To solve the equation, we need to find the value of x that satisfies the equation:

18 / (x + 1.5) = 1.5 / (x - 1.5) + 30

To make solving easier, let's first get rid of the fractions by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. The LCM of (x + 1.5) and (x - 1.5) is (x + 1.5)(x - 1.5).

18 * (x - 1.5) = 1.5 * (x + 1.5) + 30 * (x + 1.5)(x - 1.5)

Now, distribute the values on the right side of the equation:

18x - 27 = 1.5x + 2.25 + 30x^2 - 30(1.5)^2

Simplify the equation:

18x - 27 = 1.5x + 2.25 + 30x^2 - 30(2.25)

18x - 27 = 1.5x + 2.25 + 30x^2 - 67.5

Now, move all the terms to one side of the equation to form a quadratic equation:

30x^2 + 1.5x - 27 + 67.5 - 2.25 = 0

30x^2 + 1.5x + 38.25 = 0

Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 30, b = 1.5, and c = 38.25. To solve this equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Substitute the values:

x = (-1.5 ± √(1.5^2 - 4 * 30 * 38.25)) / 2 * 30

x = (-1.5 ± √(2.25 - 4590)) / 60

x = (-1.5 ± √(-4587.75)) / 60

Since the square root of a negative number is not a real number, there are no real solutions for x that satisfy the equation. Therefore, the equation has no real solutions.