tìm x biết : a,(5x-3) chia hết cho (x+3) b,(5x+3) chia hết cho (2x-7) giúp mk với , mk đang cần gấp 15/10/2021 Bởi Remi tìm x biết : a,(5x-3) chia hết cho (x+3) b,(5x+3) chia hết cho (2x-7) giúp mk với , mk đang cần gấp
a, $5x-3\vdots x+3$ $\Leftrightarrow 5x+15-18\vdots x+3$ $\Leftrightarrow 5(x+3)-18\vdots x+3$ $\Rightarrow -18\vdots x+3$ $\Rightarrow x+3\in Ư(-18)= \{ \pm 1; \pm 2; \pm 3; \pm 6;\pm 9; \pm 18\}$ $\Rightarrow x\in \{-2;-4;-1;-5;0;-6;3;-9;6;-12;15;-21\}$ b, $5x+3\vdots 2x-7$ $\Leftrightarrow 10x+6\vdots 2x-7$ $\Leftrightarrow 10x-35+41\vdots 2x-7$ $\Leftrightarrow 5(2x-7)+41\vdots 2x-7$ $\Leftrightarrow 41\vdots 2x-7$ $\Rightarrow 2x-7\in Ư(41)= \{ \pm 1; \pm 41\}$ $\Rightarrow x\in \{ 4;3;24;-17\}$ Bình luận
Đáp án: Giải thích các bước giải: $5x-3$ $\vdots$ $x+3$ $⇒5x+15-15-3$ $\vdots$ $x+3$ $⇒5.(x+3)-18$ $\vdots$ $x+3$ $⇒18$ $\vdots$ $x+3$ $⇒x+3∈${$18;9;6;3;2;1;-1;-2;-3;-6;-9;-18$} $⇒x∈${$15;6;3;0;-1;-2;-4;-5;6;-9;-12;-21$} $ $ $ $ $5x+3$ $\vdots$ $2x-7$ $⇒10x+6$ $\vdots$ $2x-7$ $⇒10x-35+35+6$ $\vdots$ $2x-7$ $⇒5.(2x-7)+41$ $\vdots$ $2x-7$ $⇒41$ $\vdots$ $2x-7$ $⇒2x-7∈${$41;1;-1;-41$} $⇒x∈${$24;4;3;-17$} Bình luận
a,
$5x-3\vdots x+3$
$\Leftrightarrow 5x+15-18\vdots x+3$
$\Leftrightarrow 5(x+3)-18\vdots x+3$
$\Rightarrow -18\vdots x+3$
$\Rightarrow x+3\in Ư(-18)= \{ \pm 1; \pm 2; \pm 3; \pm 6;\pm 9; \pm 18\}$
$\Rightarrow x\in \{-2;-4;-1;-5;0;-6;3;-9;6;-12;15;-21\}$
b,
$5x+3\vdots 2x-7$
$\Leftrightarrow 10x+6\vdots 2x-7$
$\Leftrightarrow 10x-35+41\vdots 2x-7$
$\Leftrightarrow 5(2x-7)+41\vdots 2x-7$
$\Leftrightarrow 41\vdots 2x-7$
$\Rightarrow 2x-7\in Ư(41)= \{ \pm 1; \pm 41\}$
$\Rightarrow x\in \{ 4;3;24;-17\}$
Đáp án:
Giải thích các bước giải:
$5x-3$ $\vdots$ $x+3$
$⇒5x+15-15-3$ $\vdots$ $x+3$
$⇒5.(x+3)-18$ $\vdots$ $x+3$
$⇒18$ $\vdots$ $x+3$
$⇒x+3∈${$18;9;6;3;2;1;-1;-2;-3;-6;-9;-18$}
$⇒x∈${$15;6;3;0;-1;-2;-4;-5;6;-9;-12;-21$}
$ $
$ $
$5x+3$ $\vdots$ $2x-7$
$⇒10x+6$ $\vdots$ $2x-7$
$⇒10x-35+35+6$ $\vdots$ $2x-7$
$⇒5.(2x-7)+41$ $\vdots$ $2x-7$
$⇒41$ $\vdots$ $2x-7$
$⇒2x-7∈${$41;1;-1;-41$}
$⇒x∈${$24;4;3;-17$}