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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) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
Dùng tạm ngoặc này nhé:
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\-\frac{1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{7}x=\frac{2}{7}\\-\frac{1}{5}x=-\frac{3}{5}\\\frac{1}{3}x=-\frac{4}{3}\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}}\)
\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right)x=\frac{16}{5}\)
\(\Rightarrow\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}.x=\frac{16}{5}-\frac{8}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=4\)
vậy x=4
a) (2x+3)-(5x-17)=9
2x+3-5x+17=9
-3x+20=9
-3x=9-20=-11
=>x=11/3
b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)
=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)
=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)
\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)
\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)
x=107/315:2/5=107/126
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
\(a,\left(x\cdot6\frac{2}{7}+\frac{3}{7}\right)\cdot2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x\cdot\frac{44}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(\frac{44x}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{44x+3}{7}\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{11\cdot\left(44x+3\right)}{5\cdot7}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{484x+33}{35}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{484x+33}{35}-\frac{15}{35}=-2\)
\(\Rightarrow\frac{484x+33-15}{35}=-2\)
\(\Rightarrow\frac{484x+18}{35}=-2\)
\(\Rightarrow\frac{484x+18}{35}=\frac{-70}{35}\)
\(\Rightarrow484x+18=\left(-70\right)\)
\(\Rightarrow484x=\left(-70\right)-18\)
\(\Rightarrow484x=-88\)
\(\Rightarrow x=-\frac{88}{484}=-\frac{2}{11}\)
\(b,x\cdot3\frac{1}{4}+\left(-\frac{7}{6}\right)\cdot x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\cdot\frac{13}{4}+\left(-\frac{7}{6}\right)\cdot x-\frac{5}{3}=\frac{5}{12}\)
\(\Rightarrow\frac{13x}{4}+\left(-\frac{7x}{6}\right)-\frac{5}{3}=\frac{5}{12}\)
\(\Rightarrow\frac{13x\cdot3}{12}+\left(-\frac{7x\cdot2}{12}\right)-\frac{5\cdot4}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{39x}{12}+\left(-\frac{14x}{12}\right)-\frac{20}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{39x-14x-20}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{25x-20}{12}=\frac{5}{12}\)
\(\Rightarrow25x-20=5\)
\(\Rightarrow25x=20+5\)
\(\Rightarrow25x=25\)
\(\Rightarrow x=1\)
\(\frac{7}{2}-\left(x-\frac{7}{5}\right)=\frac{3}{5}\)
\(\frac{7}{2}-x+\frac{7}{5}=\frac{3}{5}\)
\(\frac{7}{2}-x=-\frac{4}{5}\)
\(x=\frac{43}{10}\)
Bài làm
\(\frac{7}{2}-\left(x-\frac{7}{5}\right)=\frac{3}{5}\)
\(\Rightarrow x-\frac{7}{5}=\frac{7}{2}-\frac{3}{5}\)
\(\Rightarrow x-\frac{7}{5}=\frac{35}{10}-\frac{6}{10}\)
\(\Rightarrow x-\frac{7}{5}=\frac{29}{10}\)
\(\Rightarrow x=\frac{29}{10}+\frac{7}{5}\)
\(\Rightarrow x=\frac{29}{10}+\frac{14}{10}\)
\(\Rightarrow x=\frac{43}{10}\)
Vậy \(x=\frac{43}{10}\)