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LT
0
1
25 tháng 4 2016
a) x- 25%x= 1/2<=>x-1/4x=1/2<=>3/4x=1/2<=>x=1/2:3/4=2/3
b)\(\left(50\%x+2\frac{1}{4}\right).\frac{-2}{3}=\frac{17}{6}\Leftrightarrow-\frac{1}{3}x-\frac{3}{2}=\frac{17}{6}\Leftrightarrow x=-13\)
c)\(\left(3\frac{x}{7}+1\right):\left(-4\right)=\frac{-1}{28}\Leftrightarrow\frac{28+x}{7}:\left(-4\right)=\frac{-1}{28}\Leftrightarrow\frac{28+x}{-28}=\frac{1}{-28}\Leftrightarrow28+x=1\Leftrightarrow x=-27\)
d)\(\left(1\frac{1}{3}-25\%-\frac{5}{12}\right)-2x=1,6:\frac{3}{5}\Leftrightarrow\frac{2}{3}-2x=\frac{8}{3}\Leftrightarrow x=-1\)
25 tháng 4 2016
a)x-25%x=1/2
x-1/4x=1/2
x.(1-1/4)=1/2
x.3/4=1/2
x=1/2:3/4
x=2/3
LN
9 tháng 4 2017
(4/3-1/4.X-5/12)-2.X=8/3
(16-3.x-5)-24.X=32
16-3X-5-24X=32
11-27X=32
x=-7/9
9 tháng 4 2017
\(\left(\frac{3}{2}-\frac{1}{4}.x-\frac{5}{12}\right)-2.x=\frac{8}{5}:\frac{3}{5}\)
\(\left(\frac{13}{12}-\frac{1}{4}.x\right)-2.x=\frac{8}{3}\)
\(\frac{13}{12}-x.\left(\frac{1}{4}-2\right)=\frac{8}{3}\)
\(x.\left(\frac{-1}{4}\right)=\frac{13}{12}-\frac{8}{3}=-\frac{19}{12}\)
\(x=\frac{-19}{12}:\frac{-1}{4}=\frac{19}{3}\)
MS
6 tháng 6 2016
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
6 tháng 6 2016
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
=> \(\frac{4}{3}-\frac{1}{4}-\frac{5}{12}-2x=1,6:0,6\)
=> \(\frac{2}{3}-2x=\frac{8}{3}\)
=> \(2x=\frac{8}{3}-\frac{2}{3}\)
=> \(2x=2\)
=> \(x=2:2\)
=> \(x=1\)
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