Question

14x^-9x=0; 16x^=49; 2x^-11x+12=0; x^-36x+324=0;

Answer 1

Легко обращайся все правильно))

Answer 2

It seems like you have provided a set of equations involving variable "x." However, there seem to be some typographical errors in the equations you've written, as some of them are not well-formed equations. Let's try to correct them and solve each equation:

1. 14x^2 - 9x = 0: To solve this equation, we can factor out the common term "x": x(14x - 9) = 0

Now, set each factor to zero and solve for x:

1. x = 0
2. 14x - 9 = 0 14x = 9 x = 9/14

So, the solutions are x = 0 and x = 9/14.

2. 16x^2 = 49: To solve this equation, isolate x by dividing both sides by 16: x^2 = 49/16

Now, take the square root of both sides: x = ±√(49/16) x = ±(7/4)

So, the solutions are x = 7/4 and x = -7/4.

3. 2x^2 - 11x + 12 = 0: To solve this quadratic equation, we can factor it: 2x^2 - 11x + 12 = (2x - 3)(x - 4) = 0

Set each factor to zero and solve for x:

1. 2x - 3 = 0 2x = 3 x = 3/2

2. x - 4 = 0 x = 4

So, the solutions are x = 3/2 and x = 4.

4. x^2 - 36x + 324 = 0: To solve this quadratic equation, we can factor it: x^2 - 36x + 324 = (x - 18)(x - 18) = (x - 18)^2 = 0

Set (x - 18) to zero and solve for x: x - 18 = 0 x = 18

So, the only solution is x = 18.

Please double-check your original equations for accuracy. If there are any other equations you'd like help with or if you need further clarification, feel free to ask!