giúp e 2 bài tìm x với
250:(10-x)=25 3x-2018:2=23
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\(a,250:\left(10-x\right)=25\\ \Rightarrow10-x=250:25\\ \Rightarrow10-x=10\\ \Rightarrow x=10-10=0\\ b,3x-2018:2=23\\ \Rightarrow3x-1009=23\\ \Rightarrow3x=23+1009\\ \Rightarrow3x=1032\\ \Rightarrow x=1032:3=344\\ c,\left(9x-21\right):3=2\\ \Rightarrow9x-21=2\times3\\ \Rightarrow9x-21=6\\ \Rightarrow9x=21+6\\ \Rightarrow9x=27\\ \Rightarrow x=27:9=3\)
\(d,53\left(9-x\right)=53\\ \Rightarrow9-x=53:53\\ \Rightarrow9-x=1\\ \Rightarrow x=9-1=8\)
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a) ( x - 2018 ) . 3 = 0
=> x - 2018 = 0
=> x = 2018
b) 2018 . ( 3x - 18 ) = 0
=> 3x - 18 = 0
=> 3x = 18
=> x = 6
c) 25 + ( 15 + x ) = 75
40 + x = 75
x = 35
d) 136 - 2 ( 164 - x ) = 30
2 ( 164 - x ) = 106
164 - x = 53
x = 111
e) 30 - ( 14 + 2x ) = 8
14 + 2x = 22
2x = 8
x = 4
f) 56 : ( 3x - 1 ) = 7
3x - 1 = 8
3x = 9
x = 3
\(a.\left(x-2018\right).3=0\)
\(x-2018=0\)
\(x=2018\)
~ mấy câu sau cx giống vậy nhé bạn ~
nếu bạn thấy câu nào khó thì nt cho mik nhe
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a) 400 - 5x = 200
5x = 200
x = 40
b) 250 : x + 10 = 20
250 : x = 10
x = 25
c) 96 - 3 ( x + 8 ) = 42
3 ( x + 8 ) = 54
( x + 8 ) = 54 : 3
x + 8 = 18
x = 18 - 8
x = 10
d) 36 : ( x - 5 ) = 22
36 : ( x - 5 ) = 4
x - 5 = 36 : 4
x - 5 = 9
x = 9 + 5
x = 14
e) 15 x 5 ( x - 35 ) - 525 = 0
75 ( x - 35 ) - 525 = 0
75 ( x - 35 ) = 525
x - 35 = 7
x = 7 + 35
x = 42
f) [ 3 x ( 70 - x ) + 5 ] : 2 = 46
[ 3 x ( 70 - x ) + 5 ] = 92
3 x ( 70 - x ) = 87
70 - x = 87 : 3
70 - x = 29
x = 41
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Bài 1:
Để E nguyên thì \(x+5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
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25-{14-[(2-x)-(x+13)+23]}=62-{8-[(2x-10)-(29-3x)+8]}
<=>25-14+2-x-x-13+23=62-8+2x-10-29+3x+8
<=>13-x-x-13+23=2x+3x+8-10-29
<=>-x-x+23=2x+3x+(-31)
<=>23-(x+x)=2x+2x-31
<=>23-2x=2x+3x-31
<=>23+31=2x+2x+3x
<=>54=x(2+2+3)
<=>54=x.7
=>x=54/7
Vậy x=54/7
<=>54=x.7
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a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(a,250:\left(10-x\right)=25\\ \Rightarrow10-x=10\\ \Rightarrow x=0.\\b,3x-2018:2=23\\ \Rightarrow3x-1009=23\\ \Rightarrow3x=1032\\ \Rightarrow x=344. \)
a) \(250:\left(10-x\right)=25\)
\(10-x=250:25\)
\(10-x=10\)
\(x=10-10\)
\(x=0\)
b) \(3x-2018:2=23\)
\(3x-1009=23\)
\(3x=23+1009\)
\(3x=1032\)
\(x=1032:3\)
\(x=344\)