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\(a.\left[\left(6.x-39\right).7\right].4=12\)
\(\left(6.x-39\right).7=12:4\)
\(6.x-39=3:7\)
\(6.x=\frac{3}{7}+39\)
\(x=\frac{276}{7}:6\)
\(x=\frac{46}{7}\)
( câu a )
\(b.200-\left(2.x+6\right)=4^3\)
\(200-\left(2.x+6\right)=64\)
\(2.x+6=200-64\)
\(2.x=136-6\)
\(x=130:2\)
\(x=65\)
( câu b )
+) 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, \(3-\left|2x+1\right|=\left(-5\right)\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow2x+1=\pm8\)
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
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 |
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)
b, (112005 + 112004) : 112003
= ( 112005 : 112003 ) : ( 112005 : 11 2003)
= (112005 - 2003) : ( 112004-2003)
=112 : 111 = 112-1=111=11
A)3^3. 7^7
B)4^9. 3^21
C)2^4036
D)x.x.x.x.y.y=x^4. y^2
E)a.a^3.a.a.a.b.b.b=a^7.b^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 : \(x+2=-24\Leftrightarrow x=-26\)