Question

1)45-5x^2=02)5x=2x^23)x^2+1/5=2x/3 помогиите пожалуйста ооочень нужно

Answer 1

1)-5x²=-45|:(-5) x²=9 x1=3 x2=-3 2) 5x-2x²=0 2x(2,5-x)=0 2x1=0 2,5-x2=0 x1=0 x2=2,5 3) (x²+1)*3=2x*5 3x²+3-10x=0 D=(-10)²-4*3*3=100-36=64=√64=8 x1=(-(-10)+8)/2*3=18/6 x1=3 x2=(-(-10)-8)/6=2/6 x2=⅓

Answer 2

45-5x^2=0
5x^2=-45
х^2=45/5=9
х^2=9
х=3

Answer 3

Solving the Equations

Let's solve the given equations step by step:

1) Equation 1: 45 - 5x^2 = 0

To solve this equation, we need to isolate the variable x. Here's how we can do it:

Step 1: Subtract 45 from both sides of the equation: 45 - 5x^2 - 45 = 0 - 45 -5x^2 = -45

Step 2: Divide both sides of the equation by -5 to solve for x: (-5x^2) / -5 = (-45) / -5 x^2 = 9

Step 3: Take the square root of both sides of the equation: √(x^2) = √9 x = ±3

Therefore, the solutions to the equation 45 - 5x^2 = 0 are x = 3 and x = -3.

2) Equation 2: 5x = 2x^2

To solve this equation, we need to isolate the variable x. Here's how we can do it:

Step 1: Subtract 2x^2 from both sides of the equation: 5x - 2x^2 = 0

Step 2: Factor out x from the left side of the equation: x(5 - 2x) = 0

Step 3: Apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Set each factor equal to zero and solve for x:

x = 0 5 - 2x = 0

Solving the second equation: 5 - 2x = 0 -2x = -5 x = 5/2

Therefore, the solutions to the equation 5x = 2x^2 are x = 0 and x = 5/2.

3) Equation 3: x^2 + 1/5 = 2x/3

To solve this equation, we need to isolate the variable x. Here's how we can do it:

Step 1: Multiply both sides of the equation by 15 to eliminate the fractions: 15(x^2 + 1/5) = 15(2x/3) 15x^2 + 3 = 10x

Step 2: Rearrange the equation to bring all terms to one side: 15x^2 - 10x + 3 = 0

Step 3: Factor the quadratic equation or use the quadratic formula to solve for x. However, the given equation does not factor easily, so we will use the quadratic formula:

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 15, b = -10, and c = 3. Substituting these values into the quadratic formula, we get:

x = (-(-10) ± √((-10)^2 - 4(15)(3))) / (2(15)) x = (10 ± √(100 - 180)) / 30 x = (10 ± √(-80)) / 30

Since the discriminant (√(b^2 - 4ac)) is negative, the equation has no real solutions.

Therefore, the equation x^2 + 1/5 = 2x/3 has no real solutions.

I hope this helps! Let me know if you have any further questions.