Question

1. Is this equation (in comments) no solution?
2. How would I go about solving this equation (also in comments)?

Answer 1

1. \[\frac{ x-1 }{(x+1)(3x-5)\sqrt{x+3} }=0\]
2. \[(x+1)^\frac{ -1 }{2} +\frac{ (x-1)(x+1)^\frac{ 1 }{ 2 } }{ \sqrt{x^3 +1} }=0\]

Answer 2

1.
To be equal to 0, only the numerator needs to equal 0. So solve x-1 = 0.

2.
Try to move half of the expression t the left side, an with a little algebra, it should become simpler.

Answer 3

1. Ah ok.
2. so it should look kinda like this? ( I wanted to get rid of the frational exponents: \[\frac{ (x-1)\sqrt{x+1} }{ \sqrt{x^3 -1} }=\frac{ 1 }{ x+1 }\]

Answer 4

continuing 2 I had \[(x+1)(x-1)\sqrt{x+1}=\sqrt{x^3-1}\]

Answer 5

Is that x^3-1 under the square root?

Answer 6

You also made a mistake when continuing 2.

Answer 7

yes it is. Oh i see, the (x+1) multiplies to both.

Answer 8

So keep going from there.

Answer 9

would the \[(x+1)\sqrt{x+1}\] turn into just x+1?

Answer 10

or it becomes \[x \sqrt{x+1}+\sqrt{x+1}\] which makes the equation \[(x^2-1)(x \sqrt{x+1}+\sqrt{x+1})=\sqrt{x^3+1}\]

Answer 11

I actually forgot even more, the first step should actually be \[\frac{ (x-1)\sqrt{x+1} }{ \sqrt{x^3 +1} }=-\frac{ 1 }{ \sqrt{x+1} }\]

Answer 12

I think I'm gonna make a new question so i can sort myself better.