Toán Lớp 11: giải phương trình sau: $C^1_x+C^2_x+C^3_x=\dfrac{7}{2}x$
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Toán Lớp 11: giải phương trình sau: $C^1_x+C^2_x+C^3_x=\dfrac{7}{2}x$
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C_x^1 + C_x^2 + C_x^3 = \dfrac{7}{2}x\\
\Leftrightarrow \dfrac{{x!}}{{\left( {x – 1} \right)!1!}} + \dfrac{{x!}}{{\left( {x – 2} \right)!2!}} + \dfrac{{x!}}{{\left( {x – 3} \right)!3!}} = \dfrac{7}{2}x\\
\Leftrightarrow x + \dfrac{{\left( {x – 1} \right)x}}{2} + \dfrac{{\left( {x – 2} \right)\left( {x – 1} \right)x}}{6} = \dfrac{7}{2}x\\
\Leftrightarrow x\left[ {1 + \dfrac{{\left( {x – 1} \right)}}{2} + \dfrac{{\left( {x – 2} \right)\left( {x – 1} \right)}}{6} – \dfrac{7}{2}} \right] = 0\\
\Leftrightarrow x\left[ {\dfrac{{\left( {x – 1} \right)}}{2} + \dfrac{{\left( {x – 2} \right)\left( {x – 1} \right)}}{6} – \dfrac{5}{2}} \right] = 0\\
\Leftrightarrow x\left[ {3\left( {x – 1} \right) + \left( {x – 2} \right)\left( {x – 1} \right) – 15} \right] = 0\\
\Leftrightarrow x\left[ {3x – 3 + \left( {{x^2} – 3x + 2} \right) – 15} \right] = 0\\
\Leftrightarrow x\left( {{x^2} – 16} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0(L)\\
x = 4(tm)\\
x = – 4(L)
\end{array} \right.\\
\Rightarrow x = 4
\end{array}$