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\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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1/ \(x+\dfrac{1}{2}=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}-\dfrac{1}{2}\)
\(x=\dfrac{-10}{6}-\dfrac{3}{6}\)
Vậy \(x=\dfrac{-13}{6}\)
2/\(\dfrac{1}{3}-x=\dfrac{3}{5}\)
\(-x=\dfrac{3}{5}-\dfrac{1}{3}\)
\(-x=\dfrac{9}{15}-\dfrac{5}{15}\)
\(-x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{-4}{15}\)
3/ \(3-4+x=\dfrac{7}{2}\)
\(-4+x=\dfrac{7}{2}-3\)
\(-4+x=\dfrac{7}{2}-\dfrac{6}{2}\)
\(-4+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+4\)
\(x=\dfrac{1}{2}+\dfrac{8}{2}\)
Vậy \(x=\dfrac{9}{2}\)
4/ \(x-\dfrac{4}{3}=\dfrac{-7}{9}\)
\(x=\dfrac{-7}{9}+\dfrac{4}{3}\)
\(x=\dfrac{-7}{9}+\dfrac{12}{9}\)
Vậy \(x=\dfrac{5}{9}\)
5/ \(x-\dfrac{-7}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{7}{2}\)
\(x=\dfrac{5}{6}-\dfrac{21}{6}\)
Vậy \(x=\dfrac{-16}{6}=\dfrac{-8}{3}\)
6/ \(x-\dfrac{1}{5}=\dfrac{9}{10}\)
\(x=\dfrac{9}{10}+\dfrac{1}{5}\)
\(x=\dfrac{9}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{11}{10}\)
7/ \(x+\dfrac{5}{12}=\dfrac{3}{8}\)
\(x=\dfrac{3}{8}-\dfrac{5}{12}\)
\(x=\dfrac{9}{24}-\dfrac{10}{24}\)
Vậy \(x=\dfrac{-1}{24}\)
8/ \(x+\dfrac{5}{4}=\dfrac{7}{6}\)
\(x=\dfrac{7}{6}-\dfrac{5}{4}\)
\(x=\dfrac{14}{12}-\dfrac{15}{12}\)
Vậy \(x=\dfrac{-1}{12}\)
9/ \(x-\dfrac{2}{7}=\dfrac{1}{35}\)
\(x=\dfrac{1}{35}+\dfrac{2}{7}\)
\(x=\dfrac{1}{35}+\dfrac{10}{35}\)
Vậy \(x=\dfrac{11}{35}\)
10 /\(x-\dfrac{1}{5}=\dfrac{-7}{10}\)
\(x=\dfrac{-7}{10}+\dfrac{1}{5}\)
\(x=\dfrac{-7}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
Dễ thế mà không làm được thì bạn nên xem lại nhé,một hai câu thì còn được chứ cả 10 câu thế kia rõ là ỷ lại rồi bạn ạ.Thân!
1: =72/90+65/90=137/90
2: =24/56-77/56=-53/56
3: =-7/10+4/5=1/10
4: =15/100-4/100=11/100
5: =4/6-5/6=-1/6
6: =10/40-15/40-76/40=-81/40
7: =-9/10+7/18
=-81/90+35/90=-46/90=-23/45
8: =27/90-55/90=-28/90=-14/45
9: =36/60-50/60-35/60=-49/60
10: =-4/9+5/6-3/8
=-32/72+60/72-27/72
=1/72
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)