Alan

8*(75-25) + 27 - 2*2 = 423

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Or perhaps this was meant to be \(\frac{x+1}{x-1}+\frac{2x-1}{2x+1}\)

in which case we have \(\frac{-1}{-3}+\frac{-5}{-3}\rightarrow \frac{1}{3}+\frac{5}{3}\rightarrow \frac{6}{3}\rightarrow 2\)

If so, note that it's important to use brackets to remove possible ambiguity!

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No, it's not zero!  As Melody wrote, 0/0 is undefined in mathematics (though there might be some computer languages that treat it as zero).

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The sum of squares from 1 to n is given by n(n+1)(2n+1)/6, so use this with n = 2011

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Hmm! Possibly a typo:

Oh, ok asinus, I see you have assumed the original might have been such that the 3 is not part of the exponent, which is strictly correct according to the normal hierarchy rules.  However, it's extremely unlikely that that is what is meant in this sort of question I could be wrong of course!).

Your second step is irrelevant asinus!

\((216x^9)^{2/3} \rightarrow(216^{1/3}x^{9\times1/3})^2 \rightarrow (6x^3)^2 \rightarrow 6^2x^{3\times2}\rightarrow 36x^6\)

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Here you need to set the discriminant to be less than zero.

kx^2 +(3-2k)x + k-5 = 0

(3-2k)^2 -4k(k-5) < 0

I'll leave you to work this through.

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You have found the value of q that gives two equal roots!

For two different real roots your second line should be:  4^2-4(q-2) > 0

This will result in a range of possible q values (q<6 rather than q = 6).

Just substitute the values given into the equation to see if they work.

(You should find C as the result).