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a) ( x- 55 ) . 17 = 0
x - 55 = 0 : 17
x - 55 = 0
x = 0 + 55
x = 55
Vậy x = 55
b) 25 . ( x - 75 ) = 25
x - 75 = 25 : 25
x - 75 = 1
x = 1 + 75
x = 76
Vậy x = 76
c) ( x - 5 ) ( x - 7 ) = 0
=> x - 5 = 0 hoặc x - 7 = 0
x = 0 + 5 x = 0 + 7
x = 5 x = 7
Vậy \(x\in\left\{5;7\right\}\)
d) ( x -11 ) ( x + 17 ) = 0
=> x - 11 = 0 hoặc x + 17 = 0
x = 0 + 11 x = 0 - 17
x = 11 x = -17
Vậy \(x\in\left\{11;-17\right\}\)
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\(\left(x+17\right)\left(25-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+17=0\\25-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-17\\x=25\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x+17=0\\25-x=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0-17\\x=25-0\end{cases}}\)
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b ) - 25 + ( - 16 + x ) = 0
( - 16 + x ) = 0 - ( - 25 )
- 16 + x = 25
x = 25 - ( - 16 )
x = 41
Vậy x = 41
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a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
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a ) 28 + x = − 59 x = − 59 − 28 x = − 87
b ) x − 4 x 2 − 25 = 0 ⇔ x − 4 = 0 x 2 − 25 = 0 ⇔ x = 4 x 2 = 25 ⇔ x = 4 x = ± 5
x + 2 = 17 ⇒ x + 2 = 17 x + 2 = − 17 ⇒ x = 15 x = − 19
d ) − 19 + 4 ( 2 − x ) = 25 − x − 19 + 8 − 4 x = 25 − x 4 x − x = − 19 + 8 − 25 3 x = − 36 = > x = − 12
![](https://rs.olm.vn/images/avt/0.png?1311)
25+(x+17)=0
x+17 =0-25
X+17 =-25
X =-25-17
X =-42
Vậy X=-42
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-----Huongdan------\)
\(Taco:\)
\(|x+25|,|17-y|\ge0\forall x,y\)
\(Mà:|x+25|+|17-y|=0\Rightarrow\hept{\begin{cases}x+25=0\Rightarrow x=-25\\17-y=0\Rightarrow y=17\end{cases}}\)
Vì \(\left|x+25\right|\ge0vs\forall x;\left|17-y\right|\ge0vs\forall y\)
Mà \(\left|x+25\right|+\left|17-y\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+25\right|=0\\\left|17-y\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+25=0\\17-y=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-25\\y=17\end{cases}}\)
Vậy \(x=-25;y=17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+17\right)^{21}.\left(x-52\right)^{25}=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+17\right)^{21}=0\\\left(x-52\right)^{25}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x+17=0\\x-52=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-17\\x=52\end{cases}}\)
vậy \(\orbr{\begin{cases}x=-17\\x=52\end{cases}}\)
X=-42
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