Đáp án:
210
Giải thích các bước giải:
$\begin{array}{l}
{(\frac{{x + 1}}{{\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1}} - \frac{{x - 1}}{{x - \sqrt x }})^{10}}\\
= {(\frac{{(\sqrt[3]{x} + 1)(\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1)}}{{\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1}} - \frac{{(\sqrt x - 1)(\sqrt x + 1)}}{{\sqrt x .(\sqrt x - 1)}})^{10}}\\
= {(\sqrt[3]{x} + 1 - \frac{{\sqrt x + 1}}{{\sqrt x }})^{10}}\\
= {(\sqrt[3]{x} + 1 - 1 - \frac{1}{{\sqrt x }})^{10}}\\
= {(\sqrt[3]{x} - \frac{1}{{\sqrt x }})^{10}}\\
= \sum\limits_{k = 0}^{10} {{C_{10}}^k} .{(\sqrt[3]{x})^{10 - k}}.{( - 1)^k}.{(\frac{1}{{\sqrt x }})^k}\\
= \sum\limits_{k = 0}^{10} {{C_{10}}^k} .{( - 1)^k}.{x^{\frac{{10 - k}}{3} - \frac{k}{2}}}\\
= \sum\limits_{k = 0}^{10} {{C_{10}}^k} .{( - 1)^k}.{x^{\frac{{10}}{3} - \frac{5}{6}k}}\\
so\,hang\,ko\,chua\,x\,thi\,\frac{{10}}{3} - \frac{5}{6}k = 0\, \to k = 4\\
\to so\,hang\,can\,tim\,la\,{C_{10}}^4.{( - 1)^4} = 210
\end{array}$
