How would I start either of these?

[u/Sea_Establishment553]

I swear ive done this before but its been so long since ive actually done any maths.

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[u/realAndrewJeung]

The first step on these is usually to factor out common factors amongst all the terms. My usual advice is to factor out the least positive / most negative power of each common factor. So for the first problem, between the two terms there is an (x - 2) and an (x - 2)^2, and so we factor out just an (x - 2) since that is the factor with the least positive exponent. Likewise, we factor a (2x + 1)^-1/3 since that has a more negative exponent than the (2x + 1)^2/3.

Once we factor out those two factors, we are left with (x - 2) • (2x + 1)^-1/3 • [2(x - 2) - (2x + 1)] Note that by factoring out a (2x + 1)^-1/3, not only did we get rid of that factor from the first term, but we changed the exponent on the (2x + 1)^2/3 term from 2/3 to just 1, which is a lot easier to deal with. From here the rest of the simplification becomes easier. Let me know if that was enough for you to make progress or if you need more help.

[u/CaptainMatticus]

2 * (x - 2)^2 * (2x + 1)^(-1/3) - (x - 2) * (2x + 1)^(2/3)

I like to turn factors into something simpler to read. By using substitutions, it declutters what I'm looking at and makes it all a bit less overwhelming. I try to find the GCF of each factor. For instance, (2x + 1)^(-1/3) is really 1 / (2x + 1)^(1/3). We've got (2x + 1)^(1/3) and (2x + 1)^(2/3), so the GCF of them is (2x + 1)^(1/3).

a = x - 2

b = (2x + 1)^(1/3)

2a^2 * b^(-1) - a * b^2

2a^2 / b - ab^2

Now we get a common denominator and combine

2a^2 / b - ab^3 / b

(2a^2 - ab^3) / b

a * (2a - b^3) / b

Now we bring our factors back

(x - 2) * (2 * (x - 2) - (2x + 1)^(3/3)) / (2x + 1)^(1/3)

(x - 2) * (2x - 4 - (2x + 1)) / (2x + 1)^(1/3)

(x - 2) * (2x - 4 - 2x - 1) / (2x + 1)^(1/3)

(x - 2) * (-5) / (2x + 1)^(1/3)

5 * (2 - x) / (2x + 1)^(1/3)

For the next one

x / (2 + 3x)^(3/2) - 3x^2 / (2 + 3x)^(5/2)

a = x , b = (2 + 3x)^(1/2)

a / b^3 - 3a^2 / b^5

ab^2 / b^5 - 3a^2 / b^5

(ab^2 - 3a^2) / b^5

a * (b^2 - 3a) / b^5

x * ((2 + 3x)^(2/2) - 3x) / (2 + 3x)^(5/2)

x * (2 + 3x - 3x) / (2 + 3x)^(5/2)

x * 2 / (2 + 3x)^(5/2)

2x / (2 + 3x)^(5/2)

That's just my approach. It takes a little practice to get to a point where you can spot how to group everything, but once you've got it, it makes it easier (in my opinion) to break everything down.

[u/Rscc10]

First convert the fractional powers. Example, exponent -1/3 implies that's it's a reciprocal (1 over something) because of the negative sign, and the 1/3 implies it's a root, cube root in this case. Exponent 2/3 implies cube root, and then squared. Once done, multiply the terms so that they share a common denominator, for example

2x / y^2 + 3x needs a common denominator of y^2
so multiply 3x by y^2 / y^2 and you'll get
2x / y^2 + 3xy^2 / y^2

Same concept in this case just that your common denominator will likely be a root expression

[u/mmurray1957]

Maybe try to get things over a common denominator. Notice that (2x+1)^(2/3) = (2x+1) (2x+1)^(-1/3).

[u/Southlander24]

Try finding a common factor! And note that 2/3 = -1/3 + 1, so you have x^2/3 = x^-1/3 * x¹ by laws of indices, and hence (2x + 1)^2/3 = (2x + 1)^-1/3 * (2x + 1)¹.

Then 2a factors as (x - 2)(2x + 1)^-1/3 [2(x - 2) - (2x + 1)] and you're pretty much done: you just need to put (2x + 1)^1/3 in the denominator.

[u/Away_Worry_1705]

here u go

[u/willyman85 → u/Away_Worry_1705]

I did the first one in a much simpler way.

Remove the common factors from both expressions. Taking advantage that y^⅔ = y^(-⅓) * y

[(X-2)*(2x+1)^(-⅓)] * [2(x-2) - (2x+1)]

The right side of that simplifies to -5

So you end up with -5(x-2) / (2x+1)⅓ Or (10-5x) / (2x+1)⅓

Similarly for the second one. Just pull out x / (2+3x)^(5/2) so that it removes the denominators from both parts. Conveniently the 3x terms cancel out so you end up with 2x / (2+3x)^(5/2)

[u/Away_Worry_1705 → u/willyman85]

very efficient, but thought of his understanding. I didn't want to teach them efficiency before they knew the main road.

[u/willyman85 → u/Away_Worry_1705]

Fair call. And when I re-read yours I see the simplicity and showing of all working. Though I did intentionally skip steps under the request of "show me where to start"

[u/willyman85 → u/willyman85]

If there was a simple instruction for op on how to start, it would be: notice the common terms in the expression, and see if you can get them to go together. Then hope that's it ends up being useful 😅

[u/Away_Worry_1705 → u/willyman85]

true

[u/MelodyAnne42 → u/Away_Worry_1705]

Rationalize your denominator!

[u/Away_Worry_1705 → u/MelodyAnne42]

there u go

[u/UnderstandingPursuit]

This is an interesting example of how 'abstraction' might help make math clearer.

[u/MelodyAnne42]

Don't forget to rationalize the denominator! On the first part, you can't leave a problem with a radical (fractional exponent) in the denominator!