Problemas Resueltos - Razonamiento matemático
Problema 53
Si \( \cfrac{2(1 - t) + 5t}{ 4 } = 1 \), entonces \( \cfrac{t^2 - t}{ 3 } \) es igual a:
-
\( \frac{2}{3} \)
-
\( - \frac{2}{27} \)
-
\( \frac{1}{9} \)
-
\( - \frac{6}{27} \)
-
\(- \frac{2}{9} \)
Intenta resolverlo antes de ver la respuesta...
-
\( – \frac{2}{27} \)
En efecto:
\[
\begin{aligned}
&\frac{ 2(1 – t) + 5t }{ 4 } = 1
\\[2em]
&\hspace{2em}\Rightarrow
2(1 – t) + 5t = 4
\\[2em]
&\hspace{2em}\Rightarrow
2 – 2t + 5t = 4
\\[2em]
&\hspace{2em}\Rightarrow
3t = 2
\\[2em]
&\hspace{2em}\Rightarrow
t = \frac{2}{3}
\end{aligned}
\]
\[
\begin{aligned}
\frac{ 2(1 – t) + 5t }{ 4 } = 1
&\Rightarrow
2(1 – t) + 5t = 4
\\[2em]
&\Rightarrow
2 – 2t + 5t = 4
\\[2em]
&\Rightarrow
3t = 2
\\[2em]
&\Rightarrow
t = \frac{2}{3}
\end{aligned}
\]
Por lo tanto:
\[ \begin{aligned} \frac{t^2 – t}{3} &= \frac{ \left( \frac{2}{3} \right)^2 – \frac{2}{3} }{3} \\[2em] &= \frac{ \frac{4}{9} – \frac{2}{3} }{3} \\[2em] &= \frac{ \frac{ 4 – 6 }{9} }{3} \\[2em] &= \frac{ \frac{-2}{9} }{3} \\[2em] &= \frac{-2}{27} = \boldsymbol{ -\frac{2}{27} } \end{aligned} \]