Question

How do you simplify the following expressions: `(x)/(sqrt(x+5)-sqrt(5))`? `(1/x)/(x-1/2)`? `x^4+11x^2-80`? `e^x(2x+1)^3+e^2x(2x+1)^2`? `x^3-xy^2+x^2y-y^3`?

Answer 1

You have posted too many questions on the one page. Please allocate one solution request page to each question.

`sqrt(x+5)+sqrt(5)`

Explanation:

Solution to the first one.

Given:`" "x/(sqrt(x+5)-sqrt(5))`

'.......................................................................................................................
Known that if you have `a^2-b^2` then it is the same as `(a-b)(a+b)`
We have the equivalent of `a-b`
,.........................................................................................................................

Multiply by 1 but in the form of

`1=(sqrt(x+5)+sqrt(5))/(sqrt(x+5)+sqrt(5)) larr" equivalent to the examples "(a+b)` but
`" "` as `(a+b)/(a+b)`

This will result in squaring the square roots in the denominator

`x/(sqrt(x+5)-sqrt(5))xx(sqrt(x+5)+sqrt(5))/(sqrt(x+5)+sqrt(5))`

`(x(sqrt(x+5)+sqrt(5)))/((x+5)-(5)) = sqrt(x+5)+sqrt(5)`

Answer 2

I will factorize `x^4 + 11x^2 - 80` for you.

Explanation:

Problems like this are often meant to be factored as `(x^2 + a)(x^2 +b)`, `a and b` being integers.

We use the same strategy to factor trinomials of this type as we do to factor those of the form `y = ax^2 + bx + c`, namely finding two numbers that multiply to `ac` and that add to `b`. Two numbers that do this are `16` and `-5`.

Hence, `x^4 + 11x^2 - 80 = (x^2 + 16)(x^2 - 5)`

This is as far as we can go, because nothing can be factored further.

Hopefully this helps!