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a). 3. |9 - 2x| - 17 = 16
3. |9 - 2x| = 16 + 17
3. |9 - 2x| = 33
|9 - 2x| = 33 : 3
|9 - 2x| = 11
=> 9 - 2x = 11
2x = 9 - 11
2x = -2
x = - 2 : 2
x = - 1
hay 9 - 2x = - 11
2x = 9 - (- 11)
2x = 9 + 11
2x = 20
x = 20 : 2
x = 10
Vậy x = -1; x = 10
a) 3.| 9 - 2x | -17 = 16
3. | 9 - 2x | = 16 + 17 = 33
| 9 - 2x | = 33 : 3 = 11
\(\Rightarrow\)9 - 2x = 11 hoặc 9 - 2x = -11
2x = 9 - 11 2x = 9 - ( - 11 )
2x = -2 2x = 20
x = -2 : 2 x = 20 : 2
x = -1 x = 10
Bài 1:tìm x thuộc Z
a)x.(x-1)=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy: \(x=0;1\)
b)(x-3).(x+4)=0
\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Vậy: \(x=3;-4\)
c)(2x-4).(x+2)=0
\(\Leftrightarrow2\left(x-2\right).\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x=2;-2\)
d)(x+1)^2.(x-2)^2=0
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x=-1;2\)
e) x(x+1).(x+2)^2.(x+3)^3=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy: \(x=0;-1;-2;-3\)
f)(x-9)^5.(x-5)^8=0
\(\Leftrightarrow\left[\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=9\\x=5\end{matrix}\right.\)
Vậy: \(x=9;5\)
g)x(x+100)^10.(x+2000)^20.(x+300)^300=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x+100=0\\x+200=0\\x+300=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-100\\x=-200\\x=-300\end{matrix}\right.\)
Vậy: \(x=0;-100;-200;-300\)
h)(x-2)^2=0
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy: \(x=2\)
a, 3 ( x + 1 ) - 2 ( 3 x - 4 ) = - 13
=> 3x + 3 - 6x + 8 = - 13
=> 6x - 3x = 3 + 8 + 13
=> 3x = 24
=> x = 8
b, 2 ( x - 3 ) - 4 ( 2 x - 1 ) = - 20
=> 2x - 6 - 8x + 4 = - 20
=> 8x - 2x = - 6 + 4 + 20
=> 6x = 18
=> x = 3
c, 2 x ( x + 3 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
d, ( x - 1 ) ( 5 x - x ) = 0
=> \(\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)
e, ( x + 3 ) 2 ( 4 - x ) = 0
=> \(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x+3=0\\4-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
a) \(3\left(x+1\right)-2\left(3x-4\right)=-13\)
\(\Leftrightarrow3x+3-6x+8=-13\)
\(\Leftrightarrow3x-6x=-13-3-8\)
\(\Leftrightarrow-3x=-24\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
b) \(2\left(x-3\right)-4\left(2x-1\right)=-20\)
\(\Leftrightarrow2x-6-8x+4=-20\)
\(\Leftrightarrow2x-8x=-20+6-4\)
\(\Leftrightarrow-6x=-18\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
c) \(2x\left(x+3\right)=0\)
\(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
d)\(\left(x-1\right)\left(5x-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
e)\(\left(x+3\right)^2\left(4-x\right)=0\)
\(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+3=0\\-x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
a: =>3x+17=14
=>3x=-3
hay x=-1
b: =>|x+9|=-8(vô lý)
c: =>3x+2=17
=>3x=15
hay x=5
d: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)\cdot3\cdot\left(2-x\right)=0\)
hay \(x\in\left\{2;-2\right\}\)
e: =>2x+4=0
hay x=-2
f: =>2|2x-1|=34
=>|2x-1|=17
=>2x-1=17 hoặc 2x-1=-17
=>2x=18 hoặc 2x=-16
=>x=9 hoặc x=-8
Bài 1: Tìm x
a) 2x - 15 = -27
b) 2 (x + 1) – 3 = 7
c) 14 – (40 – x) = -27
d) 96 – 2(4 – 5x) = -12
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47)
f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100
h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6
k) 17/ (x + 3)3 : 3 – 1 = – 10
Bài 1: a) 2x - 15 = 27 => 2x = 27 + 15 = 42 => x = 42 : 2 = 21. b) 2(x + 1) - 3 = 7 => 2(x + 1) = 7 + 3 = 10 => x + 1 = 10 : 2 = 5 => x = 5 - 1 = 4. c) 14 - (40 - x) = -27 => 40 - x = 14 - (-27) = 41 => x = 40 - 41 = -1. d) 96 - 2(4 - 5x) = -12 => 2(4 - 5x) = 96 - (-12) = 108 => 4 - 5x = 108 : 2 = 54 => 5x = 4 - 54 = -50 => x = (-50) : 5 = -10.
a) (x-1).(x+2)=0
=> +)x-1=0=>x=1
+)x+2=0=>x=-2
vậy x thuộc {1;-2)
b) (x+4).(4-x)=0
suy ra: +) x+4=0=>x=-4
+)4-x=0=>x=4
vậy x thuộc {-4;4}
c) (x+4)(-3x+9)=0
suy ra : +) x+4= 0=>x=-4
+)-3x+9=0=>x=3
vậy x thuộc {-4;3)
d) (2x-4)(x+3)=0
suy ra : +) 2x-4=0=>x=2
+)x+3=0=>x=-3
vậy x thuộc {2;-3}
e) (x2-9).(2x+10)=0
suy ra : +) x2-9=0=>x=9/2
+) 2x+10=0=>x=-5
Vậy x thuộc {9/2;-5}
g) (4-x).x2=0
suy ra : +)4-x=0 => x=4
+) x.2=0=> x=0
Vậy x thuộc {4;0}
HT
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