2022-2023/2nd/01_Proportion_et_fractions/tpl_exercises_tech.tex
Bertrand Benjamin 15c6dac685
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Feat: ajoute les exercices techniques pour les calculs de fractions
2022-09-01 18:17:22 +02:00

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\begin{exercise}[subtitle={Multiplication de fractions}, step={3}, origin={Création}, topics={ Proportion et fractions }, tags={ Statistiques, Fractions }, mode={\searchMode}]
Faire les calculs suivants
\begin{multicols}{4}
\begin{enumerate}[label={\Alph*=}]
%- set A = random_expression("{a} / {b} + {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"rejected":[-1, 0, 1]})
\item $\Var{A}$
%- set B = random_expression("{a} / {b} + {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"rejected":[-1, 0, 1]})
\item $\Var{B}$
%- set C = random_expression("{a} / {b} + {c} / {d*b}", ["a!=b", "c!=b", "b > 1"], global_config={"min_max":(1, 10)})
\item $\Var{C}$
%- set D = random_expression("{a} / {d*b} + {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"min_max":(1, 10)})
\item $\Var{D}$
%- set E = random_expression("{a} / {b} + {c} / {d}", ["a!=b", "c!=b", "gcd(b, d) == 1"], global_config={"min_max":(1, 10)})
\item $\Var{E}$
%- set F = random_expression("{a} / {b} + {c} / {d}", ["a!=b", "c!=b", "gcd(b, d) == 1"], global_config={"min_max":(1, 10)})
\item $\Var{F}$
\item $\dfrac{1}{a} + \dfrac{1}{2a}$
\item $\dfrac{3}{5a} + \dfrac{1}{4a}$
\end{enumerate}
\end{multicols}
\end{exercise}
\begin{solution}
\begin{enumerate}[label={\Alph*=}]
\item $\Var{A.simplify().explain() | join('=')} = \Var{A.simplify().simplified}$
\item $\Var{B.simplify().explain() | join('=')} = \Var{B.simplify().simplified}$
\item $\Var{C.simplify().explain() | join('=')} = \Var{C.simplify().simplified}$
\item $\Var{D.simplify().explain() | join('=')} = \Var{D.simplify().simplified}$
\item $\Var{E.simplify().explain() | join('=')} = \Var{E.simplify().simplified}$
\item $\Var{A.simplify().explain() | join('=')} = \Var{A.simplify().simplified}$
\item $\dfrac{1}{a} + \dfrac{1}{2a} = \dfrac{2}{2a} + \dfrac{1}{2a} = \dfrac{2+1}{2a} = \dfrac{3}{2a}$
\item $\dfrac{3}{5a} + \dfrac{1}{4a} = \dfrac{12}{20a} + \dfrac{5}{20a} = \dfrac{12+5}{2a} = \dfrac{17}{2a}$
\end{enumerate}
\end{solution}
\begin{exercise}[subtitle={Multiplication de fractions}, step={3}, origin={Création}, topics={ Proportion et fractions }, tags={ Statistiques, Fractions }, mode={\searchMode}]
Faire les calculs suivants
\begin{multicols}{4}
\begin{enumerate}[label={\Alph*=}]
%- set A = random_expression("{a} / {b} * {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"rejected":[-1, 0, 1]})
\item $\Var{A}$
%- set B = random_expression("{a} / {b} * {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"rejected":[-1, 0, 1]})
\item $B = \Var{B}$
%- set C = random_expression("{a} / {b} * {c} / {d*b}", ["a!=b", "c!=b", "b > 1"], global_config={"min_max":(1, 10)})
\item $\Var{C}$
%- set D = random_expression("{a} / {d*b} * {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"min_max":(1, 10)})
\item $\Var{D}$
%- set E = random_expression("{a} / {b} * {c} / {d}", ["a!=b", "c!=b", "gcd(b, d) == 1"], global_config={"min_max":(1, 10)})
\item $\Var{E}$
%- set F = random_expression("{a} / {b} * {c} / {d}", ["a!=b", "c!=b", "gcd(b, d) == 1"], global_config={"min_max":(1, 10)})
\item $\Var{F}$
\item $\dfrac{1}{a} * \dfrac{1}{2a}$
\item $\dfrac{3}{5a} * \dfrac{1}{4a}$
\end{enumerate}
\end{multicols}
\end{exercise}
\begin{solution}
\begin{enumerate}[label={\Alph*=}]
\item $\Var{A.simplify().explain() | join('=')} = \Var{A.simplify().simplified}$
\item $\Var{B.simplify().explain() | join('=')} = \Var{B.simplify().simplified}$
\item $\Var{C.simplify().explain() | join('=')} = \Var{C.simplify().simplified}$
\item $\Var{D.simplify().explain() | join('=')} = \Var{D.simplify().simplified}$
\item $\Var{E.simplify().explain() | join('=')} = \Var{E.simplify().simplified}$
\item $\Var{A.simplify().explain() | join('=')} = \Var{A.simplify().simplified}$
\item $\dfrac{1}{a} \times \dfrac{1}{2a} = \dfrac{1\times 1}{a\times 2a} = \dfrac{1}{2a^2}$
\item $\dfrac{3}{5a} \times \dfrac{1}{4a} = \dfrac{3\times 1}{5a\times 4a} = \dfrac{3}{20a^2}$
\end{enumerate}
\end{solution}