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n+2 E Ư(6)
mà Ư(6)={-1;1;2;-2;3;-3;6;-6}
=>nE{-3;-1;0;-4;1;-5;4;-8}
vậy........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}x=\frac{5}{4}+\frac{1}{2}x\)
\(\Rightarrow\frac{2}{3}x-\frac{1}{2}x=\frac{5}{4}\)
\(\frac{1}{6}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{1}{6}\)
x = 15/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\frac{2}{3}x+\frac{1}{5}=\frac{1}{3}\times\left(\frac{-1}{6}\right)\)
\(-\frac{2}{3}x+\frac{1}{5}=-\frac{1}{18}\)
\(-\frac{2}{3}x=-\frac{1}{18}-\frac{1}{5}\)
\(\frac{-2}{3}x=\frac{-5}{90}-\frac{18}{90}=-\frac{23}{90}\)
\(x=-\frac{23}{90}:\frac{-2}{3}\)
\(x=-\frac{23}{90}.\frac{3}{-2}=\frac{23}{60}\)
Vậy \(x=\frac{23}{60}\)
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a, (x + 2) + (x + 4) + (x + 6) + ... + (x + 50) = 750
=> x + 2 + x + 4 + x + 6 + ... + x + 50 = 750
=> (x + x + x + ... + x) + (2 + 4 + 6 + ... + 50) = 750
=> 25x + (50 + 2).25 : 2 = 750
=> 25x + 52.25 : 2 = 750
=> 25x + 650 = 750
=> 25x = 100
=> x = 4
a) ( x+x+...+x)+(2+4+6+...+50)= 750
( x*25)+ (50+2)*25:2 = 750
(x*25)+ 650 = 750
x* 25 = 750 - 650 = 100
x = 100 :25 = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) |x + 4| = 17
=> \(\orbr{\begin{cases}x+4=17\\x+4=-17\end{cases}}\)
=> \(\orbr{\begin{cases}x=13\\x=-21\end{cases}}\)
b) (7 - x) - (25 + 7) = -25
=> (7 - x) - 32 = -25
=> 7 - x = -25 + 32
=> 7 - x = 7
=> x = 7 - 7
=> x = 0
c. |x + 5| = |-7|
=> |x + 5 | = 7
=> \(\orbr{\begin{cases}x+5=7\\x+5=-7\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-12\end{cases}}\)
2) 4 . (-5)2 + 2 . (-15)
= 2. 2 . 25 + 2 . (-15)
= 2.(2 . 25 - 15)
= 2 . 35
= 70
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\) \(\left|x-3\right|+4=\left|-6\right|\)
\(\Leftrightarrow \left|x-3\right|+4=6\)
\(\Leftrightarrow\left|x-3\right|=6-4\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy \(x=1\)
\(b)\) \(7-2\cdot\left|x+1\right|=\left(-3\right)^3\)
\(\Leftrightarrow7-2\cdot\left|x+1\right|=-27\)
\(\Leftrightarrow2\cdot\left|x+1\right|=7-\left(-27\right)\)
\(\Leftrightarrow2\cdot\left|x+1\right|=34\)
\(\Leftrightarrow\left|x+1\right|=34\div2\)
\(\Leftrightarrow\left|x+1\right|=17\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=17\\x+1=-17\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=16\\x=-18\end{cases}}\)
Vậy \(x\in\left\{16;-18\right\}\)
Tìm x :
a, |x-3| + 4 = |-6|
<=> |x-3| + 4 = 6
<=> |x-3| = 6 - 4
<=> |x-3| = 2
<=> x - 3 = 2 hoặc x - 3 = -2
<=> x = 2 + 3 hoặc x = ( -2 ) + 3
<=> x = 5 hoặc x = 1
Vậy x € { 5 ; 1 }
b, 7 - 2 . |x+1| = ( - 3 )^3
<=> 7 - 2 . |x+1| = -27
<=> 2 . |x+1| = 7 + 27
<=> 2 . |x+1| = 34
<=> |x+1| = 34 : 2
<=> |x+1| = 17
<=> x + 1 = 17 hoặc x + 1 = -17
<=> x = 17 - 1 hoặc x = ( -17 ) - 1
<=> x = 16 hoặc x = -18
Vậy x € { 16 ; -18 }
![](https://rs.olm.vn/images/avt/0.png?1311)
12x+3.23=23.x-4.32
12x+3.8=8.x-4.9
12x+24=8x-36
12x-8x=36-24
4x=12
x=12:4=3
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\(|x|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}-\frac{3}{4}\)
\(\left|x\right|=\frac{20}{12}-\frac{9}{12}\)
\(\left|x\right|=\frac{11}{12}\)
\(\Rightarrow\orbr{\begin{cases}\left|x\right|=\frac{11}{12}\\\left|x\right|=\frac{-11}{12}\end{cases}}\)
\(|x|-\frac{3}{4}=\frac{5}{3}\)
\(|x|=\frac{5}{3}+\frac{3}{4}\)
\(|x|=\frac{20}{12}+\frac{9}{12}\)
\(|x|=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}\)hoặc \(x=\frac{-29}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) |x-3|+5=-7
|x+3|=-7-5
|x-3|=-12( vô lí)
Vậy không có giá trị nào của x thỏa mãn đề bài
b)||x-7|-6|=3
suy ra |x-7|-6=3 hoặc |x-7|-6 = -3
|x-7|=3+6 |x-7|=-3+6
|x-7|=9 |x-7|=3
suy ra x-7=9 hoặc x-7=-9 hoặc x-7=3 hoặc x-7=-3
TH1 ; x-7=9 TH2 : x-7=-9 TH3 : x-7=3 TH4 : x-7=-3
x=9+7 x= -9+7 x=3+7 x=-3+7
x=16 x=-2 x=10 x=4
Vậy x thuộc {10;-2;10;4}
c) |x-1|=5-x
suy ra x-1=5-x hoặc x-1=-(5-x)
x+x=1+5 x-1=-5+x
2x=6 x-x=1-5|(vô lí)
x=6:2
x=3
Vậy x=3
d) |1-x|+x+3=6
|1-x|+x=6-3
|1-x|+x=3
|1-x|=3-x
suy ra 1-x=3-x hoặc 1-x=-(3-x)
TH1 : 1-x=3-x TH2 : 1-x=(-3-x)
-x+x=-1+3 (vô lí ) 1-x=-3+x
-x-x=-1-3
-2x=-4
x= -4:(-2)
x=2
vậy x=2
7 . (2 + x) - 6 . (x - 3) = 18
=> 7 . 2 + 7x - (6x - 6 . 3) = 18
=> 14 + 7x - (6x - 18) = 18
=> 14 + 7x - 6x + 18 = 18
=> (7x - 6x) + (14 + 18) = 18
=> x + 32 = 18
=> x = 18 - 32
=> x = -14
Vậy x = -14
Ta có:
7(2 + x) - 6(x - 3) = 18
7.2 + 7x - 6x - 6.3 = 18
14 + 7x - 6x - 18 = 18
14 + x - 18 = 18
14 + x = 36
x = 36 - 14
x = 22
Vậy x = 22 là giá trị cần tìm