5x-17=3.(x-3) 2x +19chia hết x+5 x+17chia hết x+5 23/10/2021 Bởi Ximena 5x-17=3.(x-3) 2x +19chia hết x+5 x+17chia hết x+5
Đáp án: a) x=4 Giải thích các bước giải: \(\begin{array}{l}a)5x – 17 = 3x – 9\\ \to 2x = 8\\ \to x = 4\\b)2x + 19 \vdots x + 5\\ \to 2\left( {x + 5} \right) + 9 \vdots x + 5\\ \to 9 \vdots x + 5\\ \to x + 5 \in U\left( 9 \right)\\ \to \left[ \begin{array}{l}x + 5 = 9\\x + 5 = – 9\\x + 5 = 3\\x + 5 = – 3\\x + 5 = 1\\x + 5 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = 4\\x = – 15\\x = – 2\\x = – 8\\x = – 4\\x = – 6\end{array} \right.\\c)x + 17 \vdots x + 5\\ \to x + 5 + 12 \vdots x + 5\\ \to 12 \vdots x + 5\\ \to x + 5 \in U\left( {12} \right)\\ \to \left[ \begin{array}{l}x + 5 = 12\\x + 5 = – 12\\x + 5 = 6\\x + 5 = – 6\\x + 5 = 4\\x + 5 = – 4\\x + 5 = 3\\x + 5 = – 3\\x + 5 = 2\\x + 5 = – 2\\x + 5 = 1\\x + 5 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = 7\\x = – 17\\x = 1\\x = – 11\\x = – 1\\x = – 9\\x = – 2\\x = – 8\\x = – 3\\x = – 7\\x = – 4\\x = – 6\end{array} \right.\end{array}\) Bình luận
Đáp án: Giải thích các bước giải: `a)5x-17=3(x-3)` `->5x-17=3x-9` `->5x-3x=-9+17` `->2x=8` `->x=4` Vậy `x=4` `b)2x+19\vdotsx+5` `->2x+10+9\vdotsx+5` `->2(x+5)+9\vdotsx+5` `->9\vdotsx+5` `->x+5∈Ư(9)={±1;±3;±9}` Ta có bảng \begin{array}{|c|c|}\hline x+5&1&-1&3&-3&9&-9\\\hline x&-4&-6&-2&-8&4&-14\\\hline \end{array} Vậy `x∈{-4;-6;-2;-8;4;-14}` `c)x+17vdotsx+5` `->x+5+12vdotsx+5` `->12vdotsx+5` `->x+5∈Ư(12)={±1;±2;±3;±4;±6;±12}` Ta có bảng: \begin{array}{|c|c|}\hline x+5&1&-1&2&-2&3&-3&4&-4&6&-6&12&-12\\\hline x&-4&-6&-3&-7&-2&-8&-1&-9&1&-11&7&-17\\\hline \end{array} Vậy `x∈{-4;-6;-3;-7;-2;-8;-1;-9;1;-11;7;-17}` Bình luận
Đáp án:
a) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)5x – 17 = 3x – 9\\
\to 2x = 8\\
\to x = 4\\
b)2x + 19 \vdots x + 5\\
\to 2\left( {x + 5} \right) + 9 \vdots x + 5\\
\to 9 \vdots x + 5\\
\to x + 5 \in U\left( 9 \right)\\
\to \left[ \begin{array}{l}
x + 5 = 9\\
x + 5 = – 9\\
x + 5 = 3\\
x + 5 = – 3\\
x + 5 = 1\\
x + 5 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = – 15\\
x = – 2\\
x = – 8\\
x = – 4\\
x = – 6
\end{array} \right.\\
c)x + 17 \vdots x + 5\\
\to x + 5 + 12 \vdots x + 5\\
\to 12 \vdots x + 5\\
\to x + 5 \in U\left( {12} \right)\\
\to \left[ \begin{array}{l}
x + 5 = 12\\
x + 5 = – 12\\
x + 5 = 6\\
x + 5 = – 6\\
x + 5 = 4\\
x + 5 = – 4\\
x + 5 = 3\\
x + 5 = – 3\\
x + 5 = 2\\
x + 5 = – 2\\
x + 5 = 1\\
x + 5 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x = – 17\\
x = 1\\
x = – 11\\
x = – 1\\
x = – 9\\
x = – 2\\
x = – 8\\
x = – 3\\
x = – 7\\
x = – 4\\
x = – 6
\end{array} \right.
\end{array}\)
Đáp án:
Giải thích các bước giải:
`a)5x-17=3(x-3)`
`->5x-17=3x-9`
`->5x-3x=-9+17`
`->2x=8`
`->x=4`
Vậy `x=4`
`b)2x+19\vdotsx+5`
`->2x+10+9\vdotsx+5`
`->2(x+5)+9\vdotsx+5`
`->9\vdotsx+5`
`->x+5∈Ư(9)={±1;±3;±9}`
Ta có bảng
\begin{array}{|c|c|}\hline x+5&1&-1&3&-3&9&-9\\\hline x&-4&-6&-2&-8&4&-14\\\hline \end{array}
Vậy `x∈{-4;-6;-2;-8;4;-14}`
`c)x+17vdotsx+5`
`->x+5+12vdotsx+5`
`->12vdotsx+5`
`->x+5∈Ư(12)={±1;±2;±3;±4;±6;±12}`
Ta có bảng:
\begin{array}{|c|c|}\hline x+5&1&-1&2&-2&3&-3&4&-4&6&-6&12&-12\\\hline x&-4&-6&-3&-7&-2&-8&-1&-9&1&-11&7&-17\\\hline \end{array}
Vậy `x∈{-4;-6;-3;-7;-2;-8;-1;-9;1;-11;7;-17}`