Negative exponent simplify

[spacewater]

problem
\(\displaystyle 2x^2(x-1)^{1/2} - 5(x-1)^{-1/2}\)
steps
\(\displaystyle (x-1)^{-1/2} (2x{^2}-5(x-1)^{1/2--1/2})\) <--- I am confused about whether negative exponent comes first or the positive one comes first at the end. I am thinking that -1/2 should come first since it's the one with 5(x-1)
my solution that doesn't correspond to the answer sheet
\(\displaystyle \frac{2x^2-5(x-1)}{x-1^{1/2}}\) = \(\displaystyle \frac{2x^2-5x-5}{x-1^{1/2}}\)

I have no idea where I went wrong. I tried to redo it 3 times and still cant figure out. can someone teach me the steps to solve this problem? thanks

 

[Deleted member 4993]

problem
\(\displaystyle 2x^2(x-1)^{1/2} - 5(x-1)^{-1/2}\)

Rule

\(\displaystyle a^{-n} \, = \, \frac{1}{a^n}\)

so

\(\displaystyle 2x^2(x-1)^{1/2} - 5(x-1)^{-1/2}\)

\(\displaystyle = \, 2x^2(x-1)^{1/2} - \frac{5}{(x-1)^{\frac{1}{2}}}\)

\(\displaystyle = \, \frac{2x^2(x-1)^{\frac{1}{2}}\cdot (x-1)^{\frac{1}{2}}}{(x-1)^{\frac{1}{2}}}- \frac{5}{(x-1)^{\frac{1}{2}}}\)

Now continue following the steps shown in other posts....

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steps
\(\displaystyle (x-1)^{-1/2} (2x{^2}-5(x-1)^{1/2--1/2})\) <--- I am confused about whether negative exponent comes first or the positive one comes first at the end. I am thinking that -1/2 should come first since it's the one with 5(x-1)
my solution that doesn't correspond to the answer sheet
\(\displaystyle \frac{2x^2-5(x-1)}{x-1^{1/2}}\) = \(\displaystyle \frac{2x^2-5x-5}{x-1^{1/2}}\)

I have no idea where I went wrong. I tried to redo it 3 times and still cant figure out. can someone teach me the steps to solve this problem? thanks

 

[Denis]

\(\displaystyle \frac{2x^2-5(x-1)}{x-1^{1/2}}\) = \(\displaystyle \frac{2x^2-5x-5}{x-1^{1/2}}\)

Even if the left expression was correct, you'd end up wrong: 2x^2 - 5(x - 1) = 2x^2 - 5x + 5 ; NOT -5 :shock:
Plus the denominators are in need of brackets...

You seem unwilling to be careful !