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1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
+) Ta có: \(15-x+17=-\left(-6\right)+\left|-12\right|\)
\(\Leftrightarrow-x+32=6+12\)
\(\Leftrightarrow-x=18-32\)
\(\Leftrightarrow x=14\)
Vậy......
+) Ta có: \(x+20=-\left(-23\right)\)
\(\Leftrightarrow x+20=23\)
\(\Leftrightarrow x=3\)
Vậy........
+) Ta có: \(-\left|-5\right|-\left(-x\right)+4=3-\left(-25\right)\)
\(\Leftrightarrow-5+x+4=3+25\)
\(\Leftrightarrow x=29\)
Vậy.....
Chúc e hok tốt và giữ lời hứa nhá ^_^
a, \(\left|x+9\right|=12+\left(-9\right)+2\)
\(\Leftrightarrow\left|x+9\right|=12-9+2=5\Leftrightarrow\left|x+9\right|=\pm5\)
TH1 : \(x+9=5\Leftrightarrow x=-4\)
TH2 : \(x+9=-5\Leftrightarrow x=-14\)
b, \(5x-16=40+x\Leftrightarrow4x=56\Leftrightarrow x=14\)
c, \(5x-7=-21-2x\Leftrightarrow7x=-14\Leftrightarrow x=-2\)
| x + 9 | = 12 + ( - 9 ) + 2
| x + 9 | = 3+2
| x + 9 | = 5
TH1: x+9 = 5 TH2: x+9 = -5
=>x=-4 =>x=-14
Vậy x 
5x - 16 = 40 + x
5x - 16 = 40 + 1x
=>5x-1x = 40+16
4x = 56
=> x = 14
5x - 7 = - 21 - 2x
=> 5x - 2x = -21 - 7
3x = -28
=> x = -9,(3)
a, \(\left|x-3\right|-2\left(-4\right)=\left(-7\right)\left(-12\right)\)
\(\Leftrightarrow\left|x-3\right|+8=84\Leftrightarrow\left|x-3\right|=76\Leftrightarrow x-3=\pm76\)
TH1 : \(x-3=76\Leftrightarrow x=79\)
TH2 : \(x-3=-76\Leftrightarrow x=-73\)
b, \(\left|x+2\right|+3=3^3=27\Leftrightarrow\left|x+2\right|=24\Leftrightarrow x+2=\pm24\)
TH1 : \(x+2=24\Leftrightarrow x=22\)
TH2 : \(x+2=-24\Leftrightarrow x=-26\)
3, \(x-5⋮x-1\)
\(x-1-4⋮x-1\)
\(-4⋮x-1\Rightarrow x-1\inƯ\left(-4\right)=\left\{\pm1;\pm4\right\}\)
| x - 1 | 1 | -1 | 4 | -4 |
| x | 2 | 0 | 5 | -3 |
4,\(3x+5⋮2x-1\Leftrightarrow6x+10⋮2x+1\)
\(\Leftrightarrow3\left(2x+1\right)+7⋮2x+1\Leftrightarrow7⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
| 2x + 1 | 1 | -1 | 7 | -7 |
| 2x | 0 | -2 | 6 | -8 |
| x | 0 | -1 | 3 | -4 |
Ta có: \(x-5⋮x-1\)
=> \(\left(x-1\right)-4⋮x-1\)
=> \(-4⋮x-1\)
Vì \(x\in Z\Rightarrow x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng sau:
| x-1 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 2 | 0 | 3 | -1 | 5 | -3 |
Vậy \(x\in\left\{2;0;3;-1;5;-3\right\}\)
\(\frac{1}{3}\times\frac{5}{7}-\frac{7}{27}\times\frac{36}{14}\)\(=\frac{1\times5}{3\times7}-\frac{7\times36}{27\times14}\)\(=\frac{5}{21}-\frac{252}{378}=\frac{5}{21}-\frac{2}{3}=\frac{5}{21}-\frac{14}{21}\)\(=\frac{5-14}{21}=\frac{-9}{21}=\frac{-3}{7}\)
\(\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14}=\frac{5}{21}-\frac{2}{3}=-\frac{9}{21}=-\frac{3}{7}\)
tạm thời cái đề bài là tìm x vậy
tiếp tục thui
=) (6x-10)-(6x-3)\(⋮\)2x-1
=)6x-10-6x+3\(⋮\)2x-1
=) (6x-6x)-(10-3)\(⋮\)2x-1
=)0-7\(⋮\)2x-1
=)-7\(⋮\)2x-1=)2x-1\(\in\)Ư(-7)={-7;-1;1;7}
=)2x\(\in\){-6;0;2;8}
=)x\(\in\){-3;0;1;4}
a)\(\frac{18}{-5}.\frac{-5}{6}=\frac{18}{6}=3\)
c)\(\frac{115}{30}\div\frac{-5}{6}=\frac{115}{30}.\frac{6}{-5}=\frac{-23}{5}\)
d)\(-3x+\frac{1}{2}=3\)
\(-3x=3-\frac{1}{2}\)
\(-3x=\frac{5}{2}\)
\(x=\frac{5}{2}\div3\)
\(x=\frac{5}{6}\)
a)= 18/-5 . -5/6 , bỏ -5=3
c)= 115/30 : 5/6 = 23/6 :5/6 =23/6 . 6/5 bỏ 6 = 23/5
d) 3.x +1/2 = 3:
3.x = 3-1/2 =5/2
x =5/2 : 3 = 5/6
nhanh con mẹ mày
a, \(3-\left|2x+1\right|=\left(-5\right)\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow2x+1=\pm8\)
TH1 : \(2x+1=8\Leftrightarrow x=\frac{7}{2}\)
TH2 : \(2x+1=-8\Leftrightarrow x=-\frac{9}{2}\)
b, \(12+\left|3-x\right|=9\Leftrightarrow\left|3-x\right|=3\Leftrightarrow3-x=\pm3\)
TH1 : \(3-x=3\Leftrightarrow x=0\)
TH2 : \(3-x=-3\Leftrightarrow x=6\)
a,3- | 2x + 1 | = ( - 5 )
| 2x + 1 | = 8
TH1: 2x+1=8
2x=7
x=7/2
TH2:2x+1=-8
2x=-9
x=-9/2