Đáp án:
$\begin{array}{l}
10)S = {\left( {\dfrac{{ - 1}}{7}} \right)^0} + {\left( {\dfrac{{ - 1}}{7}} \right)^1} + {\left( {\dfrac{{ - 1}}{7}} \right)^2} + .. + {\left( {\dfrac{{ - 1}}{7}} \right)^{2007}}\\
\Rightarrow \left( {\dfrac{{ - 1}}{7}} \right).S = {\left( {\dfrac{{ - 1}}{7}} \right)^1} + {\left( {\dfrac{{ - 1}}{7}} \right)^2} + {\left( {\dfrac{{ - 1}}{7}} \right)^3} + ... + {\left( {\dfrac{{ - 1}}{7}} \right)^{2008}}\\
\Rightarrow \left( {\dfrac{{ - 1}}{7}} \right).S - S = {\left( {\dfrac{{ - 1}}{7}} \right)^{2008}} - {\left( {\dfrac{{ - 1}}{7}} \right)^0}\\
\Rightarrow - \dfrac{8}{7}.S = \dfrac{1}{{{7^{2008}}}} - 1\\
\Rightarrow S = \dfrac{7}{8}.\left( {1 - \dfrac{1}{{{7^{2008}}}}} \right)\\
\Rightarrow S = \dfrac{7}{8} - \dfrac{1}{{{{8.7}^{2007}}}}\\
11)a){\left( {x - 1} \right)^5} = - 243\\
\Rightarrow {\left( {x - 1} \right)^5} = {\left( { - 3} \right)^5}\\
\Rightarrow x - 1 = - 3\\
\Rightarrow x = - 3 + 1\\
\Rightarrow x = - 2\\
Vậy\,x = - 2\\
b)\dfrac{{x + 2}}{{11}} + \dfrac{{x + 2}}{{12}} + \dfrac{{x + 2}}{{13}} = \dfrac{{x + 2}}{{14}} + \dfrac{{x + 2}}{{15}}\\
\Rightarrow \left( {x + 2} \right).\left( {\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} - \dfrac{1}{{14}} - \dfrac{1}{{15}}} \right) = 0\\
\Rightarrow x + 2 = 0\\
\Rightarrow x = - 2\\
Vậy\,x = - 2\\
c)x - 2\sqrt x = 0\left( {x \ge 0} \right)\\
\Rightarrow \sqrt x .\left( {\sqrt x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 0;x = 4
\end{array}$
