Đáp án + Giải thích các bước giải:
Bài `7` :
`1)` `(-6-2)(-6+2)=36+(-12)+12-4=36-12+12-4=32`
`2)` `(7*3-3):(-6)=(21-3):(-6)=18:(-6)=-3`
`3)` `(-5+9)*(-4)=4*(-4)=-16`
`4)` `72:(-6*2+4)=72:(-12+4)=72:(-8)=-9`
`5)` `-3*7 - 4*(-5)+1=-21+20+1=0`
`6)` `18-10:(+2)-7=18-10:2-7=18-5-7=6`
`7)` `15:(-5)(-3)-8=15:15-8=1-8=-7`
`8)` `(6*8-10:5)+3*(-7)=(48-2)+(-21)=46-21=25`
Bài `8` :
`1)` `x- 2 = -6 =>x=-6+2=>x=-4`
Vậy `x = -4`
`2)` `-5x-(-3)=13 =>-5x+3=13=>-5x=10=>x=-2`
Vậy `x = -2`
`3)` `15-(x-7)=-21=>x-7=15-(-21)` $\\$ `=>x-7=15+21` $\\$ `=>x-7=36` $\\$ `=>x=36+7=43`
Vậy `x = 43`
`4)` `3x + 17 = 2=>3x=2-17=>3x=-15=>x=-5`
Vậy `x = -5`
`5)` `45-(x-9)=-35` $\\$ `=>x-9=45-(-35)` $\\$ `=>x-9=45+35` $\\$ `=>x-9=80` $\\$ `=>x=89`
Vậy `x = 89`
`6)` `(-5)+x=15=>x=15-(-5)=>x=15+5=>x=20`
Vậy `x = 20`
`7)` `2x-(-17)=15`$\\$ `=>2x+17=15` $\\$ `=>2x=-2` $\\$ `=>x=-1`
Vậy `x = -1`
`8)` `|x-2|=3=>`\(\left[ \begin{array}{l}x-2=3\\x-2=-3\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.\)
Vậy `x in {5;-1}`
Bài `9` :
`1)` `|x-3|-7=13=>|x-3|=20` $\\$ `=>`\(\left[ \begin{array}{l}x-3=20\\x-3=-20\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=23\\x=-17\end{array} \right.\)
Vậy `x in {23;-17}`
`2)` `72-3|x+1|=9` $\\$ `=>3|x+1|=63` $\\$ `=>|x+1|=21 =>`\(\left[ \begin{array}{l}x+1=21\\x+1=-21\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=20\\x=-22\end{array} \right.\)
Vậy `x in {20;-22}`
`3)` `3|x-1|-5=7=>3|x-1|=12` $\\$ `=>|x-1|=4=>`\(\left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\)
Vậy `x in {5;-3}`
`4)` `-12(x-5)+7(3-x)=5` $\\$ `=>-12x+60+21-7x=5` $\\$ `=>-19x=5-21-60` $\\$ `=>-19x=-76` $\\$ `=>x=4`
Vậy `x = 4`
`5)` `x(x+7)=0=>` \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
Vậy `x in {0;-7}`
`6)` `(x-2)(x+4)=0=>`\(\left[ \begin{array}{l}x-2=0\\x+4=0\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
Vậy `x in {2;-4}`
`7)` `(x-2)(x+15)=0=>`\(\left[ \begin{array}{l}x-2=0\\x+15=0\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=2\\x=-15\end{array} \right.\)
Vậy `x in {2;-15}`
`8)` `(x+12)(x-3)=0=>`\(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
Vậy `x in {-12;3}`
`9)` `(-x+5)(3-x)=0=>`\(\left[ \begin{array}{l}-x+5=0\\3-x=0\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy `x in{5;3}`
`10)` `x(2+x)(7-x)=0=>`\(\left[ \begin{array}{l}x=0\\2+x=0\\7-x=0\end{array} \right.\) $\\$ `=>`\(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy `x in {0;-2;7}`
