Binomial‑Theorem Coefficient Technique

Solution (60‑second route)

Use the Binomial Theorem:

$$ (2x^{2}-\tfrac1x)^{7} =\sum_{k=0}^{7}\binom{7}{k} \Bigl(2x^{2}\Bigr)^{k}!\Bigl(-\tfrac1x\Bigr)^{7-k}. $$

The general term is

$$ T_k =\binom{7}{k},2^{,k}(-1)^{,7-k}, x^{,2k-(7-k)} =\binom{7}{k},2^{,k}(-1)^{,7-k}, x^{,3k-7}. $$

We want the exponent 3k−73k-7 to equal 22:

$$ 3k-7 = 2 ;\Longrightarrow; 3k = 9 ;\Longrightarrow; k = 3. $$

Plug k=3k=3 into the coefficient part:

$$ \binom{7}{3},2^{3}(-1)^{,4} =35 \times 8 \times 1 = 280. $$

Hence the coefficient of $x^{2}$ is $\boxed{280}$.