Tìm số x , biết
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=-\dfrac{53}{10}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
d) \(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
f) \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
\(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(\left(\dfrac{2}{3}-\dfrac{3}{2}\right)x=\dfrac{5}{12}\)
\(\dfrac{-5}{6}.x=\dfrac{5}{12}\)
-> x = \(\dfrac{-1}{2}\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\\ < =>x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\\ < =>-\dfrac{5}{6}x=\dfrac{5}{12}\\ =>x=\dfrac{\dfrac{5}{12}}{-\dfrac{5}{6}}=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=\dfrac{-53}{10}\\ < =>\dfrac{2}{5}+\dfrac{9}{5}x-\dfrac{111}{50}=-\dfrac{53}{10}\\ < =>\dfrac{9}{5}x=-\dfrac{2}{5}+\dfrac{111}{50}-\dfrac{53}{10}\\ < =>\dfrac{9}{5}x=-\dfrac{87}{25}\\ =>x=\dfrac{\dfrac{-87}{25}}{\dfrac{9}{5}}=-\dfrac{29}{15}=-1\dfrac{14}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{\dfrac{7}{9}}{2}+\dfrac{\dfrac{7}{9}}{\dfrac{3}{4}x}+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{7}{18}+\dfrac{28}{27}x+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{28}{27}x=-\dfrac{7}{18}-\dfrac{5}{9}+\dfrac{23}{27}\\ < =>\dfrac{28}{27}x=\dfrac{1}{54}\\ =>x=\dfrac{\dfrac{1}{54}}{\dfrac{28}{27}}=\dfrac{1}{56}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\\ < =>-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ < =>-\dfrac{2}{3}x=\dfrac{1}{10}\\ =>x=\dfrac{\dfrac{1}{10}}{-\dfrac{2}{3}}=-\dfrac{3}{20}\)
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\\ =>\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\\ =>\left|x\right|=\dfrac{29}{12}\)
Vậy: x= 29/12 hoặc x= -29/12
f) \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\\ =>\left|2x-\dfrac{1}{3}\right|=1-\dfrac{5}{6}=\dfrac{1}{6}\\ =>\left[{}\begin{matrix}\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\\\left|2x-\dfrac{1}{3}\right|=-\dfrac{1}{6}\end{matrix}\right.\\ =>x=\left[{}\begin{matrix}x=\dfrac{\dfrac{1}{6}+\dfrac{1}{3}}{2}\\x=\dfrac{\dfrac{-1}{6}+\dfrac{1}{3}}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Vậy: x= 1/4 hoặc x= 1/12
