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( X . 0,25 + 1999).2000 = (53 + 1999).2000
=> X . 0,25 +1999 = 53+1999
=> X . 0,25 = 53
=> X = 53 : 0,25 = 53 . \(\frac{1}{4}\)
=> X = \(\frac{53}{4}\)
(x.0,25+1999) . 2000=(53+1999) . 2000
(x.0,25+1999) . 2000= 2052 . 2000
(x.0,25+1999) . 2000= 4104000
x.0,25+1999 = 4104000 : 2000
x.0,25+1999 = 2052
x.0,25 = 2052-1999
x.0,25 = 53
x = 53:0.25
x = 212
Vậy x=212.
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\(\left(\dfrac{x}{7}+0,25\right)-\dfrac{1}{3}=-\dfrac{1}{28}\)
\(\dfrac{x}{7}+\dfrac{1}{4}=-\dfrac{1}{28}+\dfrac{1}{3}\)
\(\dfrac{x}{7}+\dfrac{1}{4}=\dfrac{28}{84}+\dfrac{-3}{84}\)
\(\dfrac{x}{7}+\dfrac{1}{4}=\dfrac{25}{84}\)
\(\dfrac{x}{7}=\dfrac{25}{84}-\dfrac{1}{4}\)
\(\dfrac{x}{7}=\dfrac{25}{84}-\dfrac{21}{84}\)
\(\dfrac{x}{7}=\dfrac{4}{84}\)
\(\dfrac{x}{7}=\dfrac{1}{21}\)
=> x.21=1.7
x.21=7
x=7:21
\(x=\dfrac{1}{3}\)
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\(x\cdot18:4=72\cdot0,25\)
\(x\cdot18:4=18\)
\(x\cdot4,5=18\)
\(x=18:4,5\)
\(x=4\)
\(x\times18:4=72\times0,25\)
\(\Rightarrow x\times18:4=18\)
\(\Rightarrow x\times4,5=18\)
\(\Rightarrow x=4\)
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`(x-5/12)^2 : 1 1/3= 0,25`
`=> (x-5/12)^2 : 4/3 = 0,25`
`=> (x-5/12)^2 =0,25 . 4/3`
`=> (x-5/12)^2 =1/3`
`=> (x-5/12)^2 =` \(\pm\left(\dfrac{1}{\sqrt{3}}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{12}=\dfrac{1}{\sqrt{3}}\\x-\dfrac{5}{12}=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{\sqrt{3}}+\dfrac{5}{12}\\x=-\dfrac{1}{\sqrt{3}}+\dfrac{5}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+4\sqrt{3}}{12}\\x=\dfrac{5-4\sqrt{3}}{12}\end{matrix}\right.\)
`@ Yamada-Akiro`
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[ 2x - 0,6 ] + 0,25 = 5/6
[ 2x - 0,6 ] = 5/6 - 0,25
2x - 0,6 = 7/12
2x = 7/12 + 0,6
2x = 71/60
x = 71/60 : 2
x = 71/120
\(\left(2x-0,6\right)+0,25=\frac{5}{6}\)
\(\left(2x-0,6\right)=\frac{5}{6}-0,25\)
\(2x-0,6=\frac{7}{12}\)
\(2x=\frac{7}{12}+0,6\)
\(2x=\frac{71}{60}\)
\(x=\frac{71}{60}:2\)
\(x=\frac{71}{120}\)
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(0.25-0.3x)x1/3-1/4=-5/16
(0.25-0.3x)x1/3=-1/16 (làm tắt)
0.25-0.3x=-1/16:1/3
0.25-0.3x=-3/16
0.3x=0.4375 =>x=35/24
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\(x-2014-\frac{2015}{2013}+x-2013-\frac{2015}{2014}+x-2014-\frac{2013}{2015}=3\)
\(\Rightarrow\left(x+x+x\right)+\left(-2014-2014\right)-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)
\(3x-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)
\(3x=3+2013+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2013}{2015}\)
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