Cho khai triển biểu thức {\left( {{2^{\frac{{x – 1}}{2}}} + {2^{ – \frac{x}{3}}}} \right)^n} = C_n^0{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^n} + C_n

Đáp án:

\(n=7\) và \(x=4.\)

Giải thích các bước giải:

\(\begin{array}{l}
{\left( {{2^{\frac{{x – 1}}{2}}} + {2^{ – \frac{x}{3}}}} \right)^n} = C_n^0{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^n} + C_n^1{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{n – 1}}\left( {{2^{ – \frac{x}{3}}}} \right) + … + C_n^n{\left( {{2^{ – \frac{x}{3}}}} \right)^n}\\
C_n^3 = 5C_n^1 \Leftrightarrow \frac{{n!}}{{3!\left( {n – 3} \right)!}} = 5.\frac{{n!}}{{1!\left( {n – 1} \right)!}}\\
\Leftrightarrow \frac{1}{{6\left( {n – 3} \right)!}} = \frac{5}{{\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right)!}}\\
\Leftrightarrow \left( {n – 1} \right)\left( {n – 2} \right) = 30\\
\Leftrightarrow {n^2} – 3n + 2 – 30 = 0\\
\Leftrightarrow {n^2} – 3n – 28 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
n = – 4\,\,\left( {ktm} \right)\\
n = 7\,\,\,\left( {tm} \right)
\end{array} \right..\\
So\,\,hang\,\,thu\,\,4\,\,bang\,\,20n\\
\Rightarrow C_n^3{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{n – 3}}{\left( {{2^{ – \frac{x}{3}}}} \right)^3} = 20n\\
\Leftrightarrow C_7^3{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{7 – 3}}{\left( {{2^{ – \frac{x}{3}}}} \right)^3} = 20.7\\
\Leftrightarrow {35.2^{2\left( {x – 1} \right)}}{.2^{ – x}} = 140\\
\Leftrightarrow {2^{x – 2}} = 4\\
\Leftrightarrow x – 2 = 2\\
\Leftrightarrow x = 4.
\end{array}\)