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dấu / / là giá trị tuyệt đối
a) / x + 3 / = 15
=> x + 3 = 15 hay x + 3 = -15
TH1 : x + 3 = 15
x =15 - 3
x = 12
TH2 : x + 3 = -5
x = -5 - 3
x = -8
VẬy : x thuộc { 12;-8 }
b ) /x-7/ + 13 = 25
/ x - 7 / = 25 - 13
/ x - 7 / = 12
=> x - 7 = 12 hay x - 7 = -12
TH1 : x - 7 = 12
x = 12 + 7
x =19
TH2: x - 7 = -12
x = -12 + 7
x = -5
Vậy : x thuộc { 19; -5 }
c) 3 ( x + 3 )2 = 27
( x + 3 )2 = 27 : 3
( x + 3 ) 2 = 9
x + 3 = 3 hay x + 3 = -3 ( lũy thừa bậc chẵn )
Th1 : x + 3 = 3
x = 3 -3
x = 0
TH2 : x + 3 = -3
x = -3-3
x = -6
Vậy : x thuộc { 0 ; -6 }
d) 20 - 2 ( 1-x )3 = 4
2 ( 1 - x )3 = 20 - 4
2 ( 1 - x )3 = 16
( 1 - x )3 = 16 : 2
( 1 -x )3 = 8
( 1 - x )3 = 23
=> 1 - x = 2
x = 1 - 2
x = -1
VẬy : x = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left[x+\frac{1}{3}\right]^3=-\frac{8}{27}\)
\(\Leftrightarrow\left[x+\frac{1}{3}\right]^3=\left[-\frac{2}{3}\right]^3\)
\(\Leftrightarrow x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{2}{3}-\frac{1}{3}=-1\)
Vậy x = -1
a) (x+1/3)3 = -8/27
<=> (x+1/3)3 = (-2/3)3
<=> x + 1/3 = -2/3
<=> x = -1
b) \(\frac{1}{4}x-\frac{8}{12}=1-\frac{3}{2}x\)
\(\Leftrightarrow\frac{7}{4}x=\frac{5}{3}\)
<=> x = 20/21
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải:
b) 15 - 3 . |2x - 1| = 6
=> 12 . |2x - 1| = 6
=> |2x - 1| = 6 : 12
=> |2x - 1| = 0,5
=> 2x - 1 = 0,5 hoặc 2x - 1 = -0,5
2x - 1 = 0,5 2x - 1 = -0,5
2x = 0,5 + 1 2x = -0,5 + 1
2x = 1,5 2x = 0,5
x = 1,5 : 2 = 0,75 x = 0,5 : 2 = 0,25
Vậy x = {0,75 ; 0,25}
c)2.|x| - 7 = 27
=> 2.|x| = 27 + 7
=> 2.|x| = 34
=> |x| = 34 : 2
=> |x| = 17
=> x = 17 hoặc x = -17
b , 15 -3 | 2x -1 | = 6
3 | 2x -1 | = 15 -6
3 | 2x -1 | =9
|2x -1 | = 9 : 3
|2x -1 | =3
suy ra 2x -1 thuoc { 3 ; -3 }
xet 2 truong hop
truong hop 1
2x -1 = 3
2x =3+1
2x = 4
x = 4 : 2
x = 2
c , 2| x | - 7 = 27
2 | x | =27+7
2 | x | = 34
| x | = 34 :2
| x | = 17
suy ra x thuoc { 17 ; 17 }
chuc ban hoc tot
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`2+(x+3)=7`
`\Rightarrow x+3=7-2`
`\Rightarrow x+3=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`5+(3+x)=10`
`\Rightarrow 3+x=10-5`
`\Rightarrow 3+x=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`(4+x)+1=7`
`\Rightarrow 4+x=7-1`
`\Rightarrow 4+x=6`
`\Rightarrow x=6-4`
`\Rightarrow x=2`
`(x+5)+3=9`
`\Rightarrow x+5=9-3`
`\Rightarrow x+5=6`
`\Rightarrow x=6-5`
`\Rightarrow x=1`
`(x-1)-4=7`
`\Rightarrow x-1=7+4`
`\Rightarrow x-1=11`
`\Rightarrow x=11+1`
`\Rightarrow x=12`
`4-(6-x)=1`
`\Rightarrow 6-x=4-1`
`\Rightarrow 6-x=3`
`\Rightarrow x=6-3`
`\Rightarrow x=3`
\(2+\left(x+3\right)=7\)
\(\Rightarrow2+x+3=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(5+\left(3+x\right)=10\)
\(\Rightarrow5+3+x=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=2\)
\(\left(4+x\right)+1=7\)
\(\Rightarrow4+x+1=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(\left(x+5\right)+3=9\)
\(=x+5+3=9\)
\(\Rightarrow x+8=9\)
\(\Rightarrow x=1\)
\(\left(x-1\right)-4=7\)
\(\Rightarrow x-1-4=7\)
\(\Rightarrow x-5=7\)
\(\Rightarrow x=12\)
\(4-\left(6-x\right)=1\)
\(\Rightarrow4-6-x=1\)
\(\Rightarrow-2-x=1\)
\(\Rightarrow x=-3\)
\(\text{Phần a :}\)
\(\frac{2}{3}.x=\frac{-4}{27}\)
\(\Rightarrow x=\frac{-4}{27}:\frac{2}{3}=\frac{-2}{9}\)
\(\text{Phần b :}\)
\(-1\frac{1}{3}.x=1\frac{1}{15}\)
\(\Rightarrow\frac{-4}{3}.x=\frac{16}{15}\)
\(\Rightarrow x=\frac{16}{15}:\frac{-4}{3}=\frac{-4}{5}\)