Trang chủ » Hỏi đáp » Môn Toán tìm x a, 12 là bội của 3x +1 b 3x là ước của 12 c (x+7) chia hết (x+4) 15/01/2025 tìm x a, 12 là bội của 3x +1 b 3x là ước của 12 c (x+7) chia hết (x+4)
Giải đáp: $\begin{array}{l}a)x \in \left\{ { – \dfrac{{13}}{3}; – \dfrac{7}{3};\dfrac{{ – 5}}{3};\dfrac{{ – 4}}{3}; – 1;\dfrac{{ – 2}}{3};0;\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{5}{3};\dfrac{{11}}{3}} \right\}\\b)x \in \left\{ { – 4; – 2; – \dfrac{4}{3}; – 1; – \dfrac{2}{3}; – \dfrac{1}{3};\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{4}{3};2;4} \right\}\\c)x \in \left\{ { – 7; – \dfrac{1}{4};\dfrac{1}{4};\dfrac{3}{4}} \right\}\end{array}$ Lời giải và giải thích chi tiết: $\begin{array}{l}a)12 = B\left( {3x + 1} \right)\\ \Leftrightarrow \left( {3x + 1} \right) \in Ư\left( {12} \right)\\ \Leftrightarrow \left( {3x + 1} \right) \in \left\{ \begin{array}{l} – 12; – 6; – 4; – 3; – 2; – 1;\\1;2;3;4;6;12\end{array} \right\}\\ \Leftrightarrow 3x \in \left\{ \begin{array}{l} – 13; – 7; – 5; – 4; – 3; – 2;\\0;1;2;3;5;11\end{array} \right\}\\ \Leftrightarrow x \in \left\{ { – \dfrac{{13}}{3}; – \dfrac{7}{3};\dfrac{{ – 5}}{3};\dfrac{{ – 4}}{3}; – 1;\dfrac{{ – 2}}{3};0;\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{5}{3};\dfrac{{11}}{3}} \right\}\\Vay\,x \in \left\{ { – \dfrac{{13}}{3}; – \dfrac{7}{3};\dfrac{{ – 5}}{3};\dfrac{{ – 4}}{3}; – 1;\dfrac{{ – 2}}{3};0;\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{5}{3};\dfrac{{11}}{3}} \right\}\\b)3x \in Ư\left( {12} \right)\\ \Leftrightarrow 3x \in \left\{ { – 12; – 6; – 4; – 3; – 2; – 1;1;2;3;4;6;12} \right\}\\ \Leftrightarrow x \in \left\{ { – 4; – 2; – \dfrac{4}{3}; – 1; – \dfrac{2}{3}; – \dfrac{1}{3};\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{4}{3};2;4} \right\}\\c)\left( {x + 7} \right) \vdots \left( {x + 4} \right)\\ \Leftrightarrow \left( {x + 4 + 3} \right) \vdots \left( {x + 4} \right)\\ \Leftrightarrow 3 \vdots \left( {x + 4} \right)\left( {do:\left( {x + 4} \right) \vdots \left( {x + 4} \right)} \right)\\ \Leftrightarrow \left( {x + 4} \right) \in Ư\left( 3 \right)\\ \Leftrightarrow \left( {x + 4} \right) \in \left\{ { – 3; – 1;1;3} \right\}\\ \Leftrightarrow x \in \left\{ { – 7; – \dfrac{1}{4};\dfrac{1}{4};\dfrac{3}{4}} \right\}\end{array}$ Trả lời
a)x \in \left\{ { – \dfrac{{13}}{3}; – \dfrac{7}{3};\dfrac{{ – 5}}{3};\dfrac{{ – 4}}{3}; – 1;\dfrac{{ – 2}}{3};0;\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{5}{3};\dfrac{{11}}{3}} \right\}\\
b)x \in \left\{ { – 4; – 2; – \dfrac{4}{3}; – 1; – \dfrac{2}{3}; – \dfrac{1}{3};\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{4}{3};2;4} \right\}\\
c)x \in \left\{ { – 7; – \dfrac{1}{4};\dfrac{1}{4};\dfrac{3}{4}} \right\}
\end{array}$
a)12 = B\left( {3x + 1} \right)\\
\Leftrightarrow \left( {3x + 1} \right) \in Ư\left( {12} \right)\\
\Leftrightarrow \left( {3x + 1} \right) \in \left\{ \begin{array}{l}
– 12; – 6; – 4; – 3; – 2; – 1;\\
1;2;3;4;6;12
\end{array} \right\}\\
\Leftrightarrow 3x \in \left\{ \begin{array}{l}
– 13; – 7; – 5; – 4; – 3; – 2;\\
0;1;2;3;5;11
\end{array} \right\}\\
\Leftrightarrow x \in \left\{ { – \dfrac{{13}}{3}; – \dfrac{7}{3};\dfrac{{ – 5}}{3};\dfrac{{ – 4}}{3}; – 1;\dfrac{{ – 2}}{3};0;\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{5}{3};\dfrac{{11}}{3}} \right\}\\
Vay\,x \in \left\{ { – \dfrac{{13}}{3}; – \dfrac{7}{3};\dfrac{{ – 5}}{3};\dfrac{{ – 4}}{3}; – 1;\dfrac{{ – 2}}{3};0;\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{5}{3};\dfrac{{11}}{3}} \right\}\\
b)3x \in Ư\left( {12} \right)\\
\Leftrightarrow 3x \in \left\{ { – 12; – 6; – 4; – 3; – 2; – 1;1;2;3;4;6;12} \right\}\\
\Leftrightarrow x \in \left\{ { – 4; – 2; – \dfrac{4}{3}; – 1; – \dfrac{2}{3}; – \dfrac{1}{3};\dfrac{1}{3};\dfrac{2}{3};1;\dfrac{4}{3};2;4} \right\}\\
c)\left( {x + 7} \right) \vdots \left( {x + 4} \right)\\
\Leftrightarrow \left( {x + 4 + 3} \right) \vdots \left( {x + 4} \right)\\
\Leftrightarrow 3 \vdots \left( {x + 4} \right)\left( {do:\left( {x + 4} \right) \vdots \left( {x + 4} \right)} \right)\\
\Leftrightarrow \left( {x + 4} \right) \in Ư\left( 3 \right)\\
\Leftrightarrow \left( {x + 4} \right) \in \left\{ { – 3; – 1;1;3} \right\}\\
\Leftrightarrow x \in \left\{ { – 7; – \dfrac{1}{4};\dfrac{1}{4};\dfrac{3}{4}} \right\}
\end{array}$