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|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
Tìm x :
a) \(2.x.\frac{-3}{4}=-\frac{5}{12}\)
\(\Rightarrow2x=-\frac{5}{12}:-\frac{3}{4}\)
\(\Rightarrow2x=\frac{5}{9}\)
\(\Rightarrow x=\frac{5}{9}:2\)
\(\Rightarrow x=\frac{5}{18}\)
Vậy : \(x=\frac{5}{18}\)
b) \(\frac{2}{3}+\frac{1}{3}.x=7\)
\(\Rightarrow\frac{1}{3}.x=7-\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}.x=\frac{19}{3}\)
\(\Rightarrow x=\frac{19}{3}:\frac{1}{3}\)
\(\Rightarrow x=19\)
Vậy : \(x=19\)
c) \(\left(4.x+\frac{1}{8}\right)=\frac{3}{10}\)
\(\Rightarrow4.x=\frac{3}{10}-\frac{1}{8}\)
\(\Rightarrow4.x=\frac{7}{40}\)
\(\Rightarrow x=\frac{7}{40}:4\)
\(\Rightarrow x=\frac{7}{160}\)
Vậy : \(x=\frac{7}{160}\)
d) \(\frac{1}{3}.x-5=1\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}.x-5=\frac{3}{2}\)
\(\Rightarrow\frac{1}{3}.x=\frac{3}{2}+5\)
\(\Rightarrow\frac{1}{3}.x=\frac{13}{2}\)
\(\Rightarrow x=\frac{13}{2}:\frac{1}{3}\)
\(\Rightarrow x=\frac{39}{2}\)
Vậy : \(x=\frac{39}{2}\)
e) \(-\frac{2}{3}.x+\frac{1}{3}=-\frac{1}{2}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{1}{2}-\frac{1}{3}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{5}{6}\)
\(\Rightarrow x=-\frac{5}{6}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=\frac{5}{4}\)
Vậy : \(x=\frac{5}{4}\)
a) 2 - ( \(5\frac{3}{8}\)x X - \(\frac{5}{24}\)) = \(\frac{5}{12}\)
\(5\frac{3}{8}\)x X - \(\frac{5}{24}\)= \(\frac{19}{12}\)
\(5\frac{3}{8}\)x X = \(\frac{43}{24}\)
X = \(\frac{1}{3}\)
b) \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x X + \(\frac{1}{6}\)) = \(\frac{22}{23}\)
\(3\frac{1}{3}\)x X + \(\frac{1}{6}\) = \(\frac{23}{18}\)
\(3\frac{1}{3}\)x X = \(\frac{10}{9}\)
X =\(\frac{1}{3}\)
C) \(\frac{4}{5}\)x X - \(\frac{1}{2}\)x X + \(\frac{3}{4}\)x X = \(\frac{7}{40}\)
( \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\)) x X = \(\frac{7}{40}\)
\(\frac{21}{20}\) x X = \(\frac{7}{40}\)
X =\(\frac{1}{6}\)
g)=>x+1/2=0
x=0-1/2
x=-1/2
hoặc 2/3-2x=0
2x=2/3-0
2x=2/3
x=2/3:2
x=1/3
nhìn @_@ hoa cả mắt đăng từng bài thôi bạn
\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
\(\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
\(\Rightarrow\frac{3}{4}-x=\orbr{\begin{cases}\frac{17}{12}\\-\frac{17}{12}\end{cases}\Rightarrow x=\orbr{\begin{cases}-\frac{2}{3}\\\frac{13}{6}\end{cases}}}\)
\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(x\left(\frac{1}{2}+\frac{1}{8}\right)=\frac{3}{4}\)
\(x.\frac{5}{8}=\frac{3}{4}\)
\(\Rightarrow x=\frac{3}{4}:\frac{5}{8}\)
\(\Rightarrow x=\frac{6}{5}\)