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a)\(\frac{5}{6}-x=-\frac{7}{12}+\frac{2}{3}\)
\(\frac{5}{6}-x=\frac{1}{12}\)
\(x=\frac{5}{6}-\frac{1}{12}\)
\(\Rightarrow x=\frac{3}{4}\)
b)\(\left(2,4x-36\right):1\frac{5}{7}=-14\)
\(\left(2,4x-36\right)=-24\)
\(2,4x=12\)
\(\Rightarrow x=5\)
c)\(\left(3\frac{1}{2}+2x\right).3\frac{2}{3}=5\frac{1}{3}\)
\(3\frac{1}{2}+2x=\frac{16}{11}\)
\(2x=-\frac{45}{22}\)
\(x=-\frac{45}{44}\)
d)\(\frac{5}{6}-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{3}{8}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{11}{24}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{11}{24}\\\frac{1}{2}x-\frac{1}{3}=-\frac{11}{24}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{19}{12}\\x=-\frac{1}{4}\end{cases}}\)
e)\(\left|\frac{1}{4}-2x\right|-\frac{3}{4}=0\)
\(\left|\frac{1}{4}-2x\right|=\frac{3}{4}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-2x=\frac{3}{4}\\\frac{1}{4}-2x=-\frac{3}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{2}\end{cases}}\)
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
\(\Rightarrow x-\frac{5}{21}=\left|1\frac{1}{6}\right|\)
\(\Rightarrow x-\frac{5}{21}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}+\frac{5}{21}=\frac{49}{42}+\frac{10}{42}=\frac{59}{42}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow x+\left|-1\frac{2}{3}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{3}{4}=-\left|-1\frac{2}{3}\right|\)
\(\Rightarrow x-\frac{3}{4}=-1\frac{2}{3}\)
\(\Rightarrow x-\frac{3}{4}=-\frac{5}{3}\)
\(\Rightarrow x=-\frac{5}{3}+\frac{3}{4}=-\frac{11}{12}\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{2}\\x-\frac{1}{3}=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}+\frac{1}{3}=\frac{17}{6}\\x=-\frac{5}{2}+\frac{1}{3}=-\frac{13}{6}\end{matrix}\right.\)
4) \(\left|x+\frac{2}{3}\right|=0\)
\(\Rightarrow x+\frac{2}{3}=0\)
\(\Rightarrow x=0-\frac{2}{3}=-\frac{2}{3}\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
\(\Rightarrow\left|x+2\right|=\frac{2}{15}\)
\(\Rightarrow\left[{}\begin{matrix}x+2=\frac{2}{15}\\x+2=-\frac{2}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{15}-2=-\frac{28}{15}\\x=-\frac{2}{15}-2=-\frac{32}{15}\end{matrix}\right.\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
\(\Rightarrow\left|x-4\right|=\frac{19}{20}\)
\(\Rightarrow\left[{}\begin{matrix}x-4=\frac{19}{20}\\x-4=-\frac{19}{20}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{19}{20}+4=\frac{99}{20}\\x=-\frac{19}{20}+4=\frac{61}{20}\end{matrix}\right.\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
Vì \(\left|x-\frac{5}{4}\right|\ge0\)
=> Không có giá trị x thỏa mãn với điều kiện trên
a) \(\frac{-2}{5}+\frac{5}{6}.x=\frac{-4}{15}\)
\(\frac{5}{6}.x=\frac{-4}{15}-\frac{-2}{5}\)
\(\frac{5}{6}.x=\frac{2}{15}\)
\(x=\frac{2}{15}:\frac{5}{6}\)
\(x=\frac{4}{25}\)
b) \(\left(x-\frac{1}{5}\right)\left(y+\frac{1}{2}\right)\left(z-3\right)=0\)
\(x-\frac{1}{5}=0\)
\(x=0+\frac{1}{5}\)
\(x=\frac{1}{5}\)