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A , 3 - ( 17 - x ) = 289 - ( 36 + 289 )
3 - 17 + x = 0 - 36
-14 + x = -36
x = -36 - ( - 14 ) = -22
B, 25 - ( x + 5 ) = -415 - ( 15 - 415 )
25 - x - 5 = 0 - 15
20 - x = -15
x = 20 - ( - 15 ) = 35
C , 34 + ( 21 - x ) = ( 3747 - 30 ) - 3746
34 + 21 - x = 1 - 30
55 - x = -29
x = 55 - (-29 ) = 74
D , -2x - ( x -17 ) = 34 - ( -x + 25 )
- 2x - x + 17 = 34 - 25 + x
- 3x + 17 = 9 + x
- 3x - x = 9 - 17
-4x = -8
x = -8 : ( - 4 )
x = 2
E , 17x + ( -16x - 37 ) = x + 43
17x - 16x -37 = x + 43
x - 37 = x + 43
-37 - 43 = x - x
- 80 = 0 ( vô lý )
G , ( x + 12 ) . (x - 3 ) = 0
\(\hept{\begin{cases}x+12=0\\x-3=0\end{cases}}\)
\(\hept{\begin{cases}x=-12\\x=3\end{cases}}\)
a, 28+2x=35-(-13)
=> 2x=35+13-28
=>2x=20
=> x=10. vậy x=10
chúc bn hok tốt k cho mik nha
1. 3x - 17 = x+3
3x - x = 3 + 17
2x = 20
x = 20 :2
x = 10
2. |x-3| - 12 = |-5|
|x-3| -12 = 5
|x-3| = 5 + 12
|x-3| = 17
=> x-3 = +17 , -17
- x - 3 = 17
x = 17 + 3
x = 20
- x - 3 = -17
x = -17 + 3
x = -14
3 . 25 - ( x -5 ) = -415 -( 15 -415)
25 - x +5 = -415 - 15 +415
25 - x +5 = ( -415 + 415) -15
25 - x +5 = 0-5 = -5
25 - x = -5-5
25 - x = -10
x = 25 - -10
x = 35
4. ( x -3 ) . ( 2x +6 ) = 0
=> x-3 = 0 hoặc 2x +6 = 0
- x - 3 = 0
x = 0 + 3 = 3
- 2x +6 = 0
2x = 0-6=-6
x = -6 : 2
x = -3
\(1;3x-17=x+3\)
\(\Leftrightarrow3x-x=3+17\)
\(\Leftrightarrow2x=20\)
\(\Rightarrow x=10\)
\(2;\left|x-3\right|-12=\left|-5\right|\)
\(\Leftrightarrow\left|x-3\right|-12=5\)
\(\Leftrightarrow\left|x-3\right|=5+12\)
\(\Leftrightarrow\left|x-3\right|=17\)
\(\Rightarrow\orbr{\begin{cases}x-3=17\\3-x=17\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=17+3\\-x=17-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=20\\x=-14\end{cases}}}\)
\(3;25-\left(x-5\right)=-415-\left(15-415\right)\)
\(\Leftrightarrow25-x+5=-415-15+415\)
\(\Leftrightarrow-x=-415-15+415-25-5\)
\(\Leftrightarrow-x=-45\)
\(\Rightarrow x=45\)
\(4;\left(x-3\right)\left(2x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
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
a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)
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