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a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)
\(-\frac{5}{6}\times x=\frac{5}{2}\)
\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)
\(x=-3\)
b.
\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)
\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)
\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=-\frac{19}{2}+\frac{37}{10}\)
\(3x=\frac{-95+37}{10}\)
\(3x=-\frac{58}{10}\)
\(3x=-\frac{29}{5}\)
\(x=-\frac{29}{5}\div3\)
\(x=-\frac{29}{5}\times\frac{1}{3}\)
\(x=-\frac{29}{15}\)
c.
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21-16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}\times\frac{4}{3}\)
\(x=\frac{5}{6}\)
d.
\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)
\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)
\(-\frac{2}{3}\times x=\frac{3-2}{10}\)
\(-\frac{2}{3}\times x=\frac{1}{10}\)
\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)
\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)
\(x=-\frac{3}{20}\)
e.
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{20+9}{12}\)
\(\left|x\right|=\frac{29}{12}\)
\(x=\pm\frac{29}{12}\)
Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)
f.
\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)
\(2x-\frac{1}{3}=\pm\frac{1}{6}\)
- \(2x-\frac{1}{3}=\frac{1}{6}\)
\(2x=\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{1+2}{6}\)
\(2x=\frac{3}{6}\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{2}\div2\)
\(x=\frac{1}{2}\times\frac{1}{2}\)
\(x=\frac{1}{4}\)
- \(2x-\frac{1}{3}=-\frac{1}{6}\)
\(2x=-\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{-1+2}{6}\)
\(2x=\frac{1}{6}\)
\(x=\frac{1}{6}\div2\)
\(x=\frac{1}{6}\times\frac{1}{2}\)
\(x=\frac{1}{12}\)
Vậy x = 1/4 hoặc x = 1/12.
Chúc bạn học tốt
Sorry nha, mik chép lộn đề
Làm lại câu a nha
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{12}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)
\(-\frac{5}{6}\times x=\frac{5}{12}\)
\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)
\(x=-\frac{1}{2}\)
Chúc bạn học tốt
a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)