Answer
$-25x^{47}$
Work Step by Step
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!},$
$n!=n(n-1)\cdot...\cdot 2\cdot 1,$
$n!=0$
--------
$...$In all the terms, exponents of a and b add to n...
In $(a+(-b))^{25} $, the second term is
$\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}(-b)^{1}=\left(\begin{array}{l}
25\\
1
\end{array}\right)a^{24}(-b)^{1}=-25a^{24}b$
So, for $(x^{2}-\displaystyle \frac{1}{x})^{25}$ the term is
$-25(x^{2})^{24}\displaystyle \cdot(\frac{1}{x})=-25x^{48-1}=$
$=-25x^{47}$
