a, 2.{x-2}+7=x-25 b,|x+5|=14+[-5] c,|x-3|=-21+37 d, |x+23|=-6 e, |x+2|+5=12

0 bình luận về “a, 2.{x-2}+7=x-25 b,|x+5|=14+[-5] c,|x-3|=-21+37 d, |x+23|=-6 e, |x+2|+5=12”

1. Đáp án:

    

   Giải thích các bước giải:

    `a,2(x-2)+7=x-25`

   `\to 2x-4+7=x-25`

   `\to 2x+3=x-25`

   `\to 2x-x=-25-3`

   `\to x=-28`

   Vậy `x=-28`

   .

   `b,|x+5|=14+(-5)`

   `\to |x+5|=9`

   `\to `\(\left[ \begin{array}{l}x+5=9\\x+5=-9\end{array} \right.\)

   `\to`\(\left[ \begin{array}{l}x=4\\x=-14\end{array} \right.\) 

   Vậy `x=4` hoặc `x=-14`

   ,

   `c,|x-3|=-21+37`

   `\to |x-3|=16`

   `\to`\(\left[ \begin{array}{l}x-3=16\\x-3=-16\end{array} \right.\)

   \(\to\left[ \begin{array}{l}x=16+3\\x=-16+3\end{array} \right.\)

   \(\to\left[ \begin{array}{l}x=19\\x=-13\end{array} \right.\) 

   Vậy `x=19` hoặc `x=-13`

   ,

   `d,|x+23|=-6`

   `\to`\(\left[ \begin{array}{l}x+23=6\\x+23=-6\end{array} \right.\)

   \(\to\left[ \begin{array}{l}x=6-23\\x=-6-23\end{array} \right.\)

   \(\to\left[ \begin{array}{l}x=-17\\x=-29\end{array} \right.\) 

   Vậy `x=-17` hoặc `x=-29

   ,

   `e,|x+12|+5=12`

   `\to|x+12|=12-5`

   `\to |x+12|=7`

   `\to`\(\left[ \begin{array}{l}x+12=7\\x+12=-7\end{array} \right.\)

   \(\to\left[ \begin{array}{l}x=7-12\\x=-7-12\end{array} \right.\)

   \(\to\left[ \begin{array}{l}x=-5\\x=-19\end{array} \right.\) 

   Vậy `x=-5` hoặc `x=-19`

   Bình luận

2. @lovemath2k9

   @xin hay nhất ạ

   $a)2.{x-2}+7=x-25$

   $2x-2.2+7=x-25$

   $2x-4+7=x-25$

   $2x+3=x-25$

   $2x-x=-25-3$

   $1x=-28$

   $x=-28:1$

   $x=-28$

   $b)|x+5|=14+(-5)$

   $|x+5|=9$

   $TH1:x+5=9$

   $x=9-5$

   $x=4$

   $TH2:x+5=-9$

   $x=-9-5$

   $x=-14$

   vậy \(\left[ \begin{array}{l}x=4\\x=-14\end{array} \right.\) 

   $c)|x-3|=-21+37$

   $|x-3|=16$

   $TH1:x-3=16$

   $x=16+3$

   $x=9$

   $TH2:x-3=-16$

   $x=-16+3$

   $x=-13$

   vậy \(\left[ \begin{array}{l}x=9\\x=-13\end{array} \right.\)  

   $d)|x+23|=-6$

   $TH1:x+23=-6$

   $x=-6-23$

   $x=-29$

   $TH2:x+23=6$

   $x=6-23$

   $x=-17$

   vậy \(\left[ \begin{array}{l}x=-29\\x=-17\end{array} \right.\) 

   $e)|x+2|+5=12$

   $|x+2|=12-5$

   $|x+2|=7$

   $TH1:x+2=7$

   $x=7-2$

   $x=5$

   $TH2:x+2=-7$

   $x=-7-2$

   $x=-9$

   vậy \(\left[ \begin{array}{l}x=5\\x=-9\end{array} \right.\) 

   Bình luận