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First, give them a common denominator by multiplying (x-3) and (3-x), which gives you -1(x-3)^2.

Then, you have to multiply the first 3 of the numerator with (3-x) and x with (x-3). This will give you x^2-6x+9 over -(x-3)^2, which will simplify to (x-3)^2 over -(x-3)^2.

The top and the bottom of the fraction will simplify and give you 1 over -1, which is simply:

-1.

Q: What is the sum of 3 over x-3 and x over 3-x?

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1-3x1=3x 1/3=x

x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)

The Answer:The sum of cubes, operation is: (a+b)(a2-ab+b2) = a3+b3So, since 27 = 33 we can reformulate to:So, x3+33 = (x + 3)(x2-3x+32) = (x + 3)(x2-3x+9).

(x-2)(x^2+3)

(X + 1)3 = X3 + 3X2 + 3X + 1

Related questions

3/x2

x3 + x2 - 3x - 3 x(x2 + x - 3) - 3

3 - 3x + x2 - x3 = (1 - x)(x2 + 3)

(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)

3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)

(x - 3)(x2 + 3x + 3)

3x over x + y

1-3x1=3x 1/3=x

x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)

No. 3X is multiple (3 x X) X3 is exponential (X x X x X)

The sum of 3x plus 2 and -4 plus 3 is 3x+1 if the equation is (3x+2)+(-4+3). If they are two separate questions, 3x+2 remains as itself while -4+3=-1.

the derivative of 3x is 3 the derivative of x cubed is 3 times x squared