Bài 1:Tìm x biết 1,X + 3 chia hết cho X-2 2, 2X + 1 chia hết cho X-5 24/10/2021 Bởi Eloise Bài 1:Tìm x biết 1,X + 3 chia hết cho X-2 2, 2X + 1 chia hết cho X-5
1/ $x+3\vdots x-2$ $→x-2+5\vdots x-2$ $→5\vdots x-2$ $→x-2\in Ư(5)=\{±1;±5\}$ $→$ Ta có bảng $\begin{array}{|c|c|c|}\hline x-2&1&-1&5&-5\\\hline x&3&1&7&-3\\\hline \quad&tm&tm&tm&tm\\\hline\end{array}$ 2/ $2x+1\vdots x-5$ $→2x-10+11\vdots x-5$ $→2(x-5)+11\vdots x-5$ $→11\vdots x-5$ $→x-5\in Ư(11)=\{±1;±11\}$ $→$ Ta có bảng $\begin{array}{|c|c|c|}\hline x-5&1&-1&11&-11\\\hline x&6&4&16&-6\\\hline \quad&tm&tm&tm&tm\\\hline\end{array}$ Bình luận
`a,` Có: `x+3 = x – 2 + 5` Vì `(x – 2) \vdots (x-2)` `⇔5 \vdots x – 2` `⇒x-2 ∈ Ư(5)={±1; ±5}` `⇒x ∈ {-3; 1; 3; 7}` `b,` Có: `2x + 1 = 2x – 10 + 11` `=2(x-5) + 11` Vì `2(x-5) \vdots (x-5)` `⇔11 \vdots x – 5` `⇒x – 5 ∈ Ư(11)={±1; ±11}` `⇒x∈{-6; 4; 6; 16}` Bình luận
1/ $x+3\vdots x-2$
$→x-2+5\vdots x-2$
$→5\vdots x-2$
$→x-2\in Ư(5)=\{±1;±5\}$
$→$ Ta có bảng
$\begin{array}{|c|c|c|}\hline x-2&1&-1&5&-5\\\hline x&3&1&7&-3\\\hline \quad&tm&tm&tm&tm\\\hline\end{array}$
2/ $2x+1\vdots x-5$
$→2x-10+11\vdots x-5$
$→2(x-5)+11\vdots x-5$
$→11\vdots x-5$
$→x-5\in Ư(11)=\{±1;±11\}$
$→$ Ta có bảng
$\begin{array}{|c|c|c|}\hline x-5&1&-1&11&-11\\\hline x&6&4&16&-6\\\hline \quad&tm&tm&tm&tm\\\hline\end{array}$
`a,`
Có: `x+3 = x – 2 + 5`
Vì `(x – 2) \vdots (x-2)`
`⇔5 \vdots x – 2`
`⇒x-2 ∈ Ư(5)={±1; ±5}`
`⇒x ∈ {-3; 1; 3; 7}`
`b,`
Có: `2x + 1 = 2x – 10 + 11`
`=2(x-5) + 11`
Vì `2(x-5) \vdots (x-5)`
`⇔11 \vdots x – 5`
`⇒x – 5 ∈ Ư(11)={±1; ±11}`
`⇒x∈{-6; 4; 6; 16}`