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6x = 511 : 59 + 31 - x
<=> 6x = 52 + 3 - x
<=> 6x = 25 + 3 - x
<=> 6x + x = 28
<=> 7x = 28
=> x = 28 : 7
=> x = 4
Vậy x = 4
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6x + x = 511 : 59 + 31
x . ( 6 + 1 ) = 52 + 3
x . 7 = 25 + 3
x . 7 = 28
x = 28 : 7
x = 4
Vậy x = 4
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6x+x = 5^2+3
x.(6+1) = 25+3
x.7=28
x=28:7
x=4
Vậy x=4
Nhớ k cho mình nhé!
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Ta có: \(6x+x=5^{11}:5^9+3^1\\ < =>7x=5^2+3^1\\ < =>7x=25+3=28\\ =>x=\dfrac{28}{7}=4\)
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Bài 1:
1; 5\(x\) + \(x\) = 39 - 311 : 39
\(x\).(5 + 1) = 39 - 32
\(x.6\) = 39 - 9
\(x.6\) = 30
\(x\) = 30 : 6
\(x\) = 5
Vậy \(x\) = 5
2; 5\(x\) + \(x\) = 150 : 2 + 3
\(x\).(5 + 1) = 75 + 3
\(x.6\) = 78
\(x\) = 78 : 6
\(x\) = 13
Vậy \(x=13\)
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câu 1:23+3x=5mu3=125-23=102:3=34.the la xong.minh lam tat
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Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
6x+x=511:59+31
7x=25+3
7x=28
x=28/7
x=4
\(\Rightarrow7x=5^{11-9}+3\)
\(\Rightarrow7x=5^2+3\)
\(\Rightarrow7x=28\)
\(\Rightarrow x=\frac{28}{7}\)