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- \(x\cdot2\frac{4}{9}=4\frac{2}{5}-1\frac{4}{7}\)
\(x\cdot\frac{22}{9}=\frac{22}{5}-\frac{11}{7}\)
\(x\cdot\frac{22}{9}=\frac{99}{35}\)
\(x=\frac{99}{35}:\frac{22}{9}\)
\(x=\frac{81}{70}\)
Vậy \(x=\frac{81}{70}\).
- \(x:10\frac{6}{5}=4\frac{6}{5}+9\frac{7}{5}\)
\(x:\frac{56}{5}=\frac{26}{5}+\frac{52}{5}\)
\(x:\frac{56}{5}=\frac{78}{5}\)
\(x=\frac{78}{5}\cdot\frac{56}{5}\)
\(x=\frac{4368}{25}\)
Vậy \(x=\frac{4368}{25}\).
- \(x:3\frac{10}{3}=9\frac{1}{9}-5\frac{1}{9}\)
\(x:\frac{19}{3}=4\)
\(x=4\cdot\frac{19}{3}\)
\(x=\frac{76}{3}\)
Vậy \(x=\frac{76}{3}\).
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
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!
a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)
\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)
\(\Leftrightarrow5x-3x+4x=35-15-48+48\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)
\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)
\(\Leftrightarrow6x-28x-18x=30+32-5-9\)
\(\Leftrightarrow-40x=48\)
\(\Leftrightarrow x=-1.2\)
\(a,-5.\left(-x+7\right)-3.\left(-x-5\right)=-4.\left(12-x\right)+48\)
\(5x-35-3x+15=-48+4x+48\)
\(2x-10=4x\)
\(2x-4x=10\)
\(-2x=10\)
\(x=-5\)
\(b,\left(-x-7\right)-5.\left(-x-3\right)=12.\left(3-x\right)\)
\(-x-7+5x+15=36-12x\)
\(4x+8=36-12x\)
\(4x+12x=36-8\)
\(16x=28\)
\(x=1,75\)
các câu còn lại tương tự nha
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 \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)
`@` `\text {Ans}`
`\downarrow`
`1)`
\(x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow` \(x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{7}{6}\)
Vậy, `x =`\(\dfrac{7}{6}\)
`2)`
\(\dfrac{3}{5}-x=\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{4}{15}\)
Vậy, `x =`\(\dfrac{4}{15}\)
`3)`
\(\dfrac{3}{4}+x=\dfrac{7}{2}\)
`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)
`\Rightarrow`\(x=\dfrac{11}{4}\)
Vậy, \(x=\dfrac{11}{4}\)
`4)`
\(x-\dfrac{4}{3}=\dfrac{7}{9}\)
`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)
`\Rightarrow`\(x=\dfrac{19}{9}\)
Vậy, `x=`\(\dfrac{19}{9}\)
`5)`
\(x-\dfrac{5}{6}=\dfrac{7}{3}\)
`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)
`\Rightarrow x =`\(\dfrac{19}{6}\)
Vậy, `x=`\(\dfrac{19}{6}\)
`6)`
\(x-\dfrac{1}{5}=\dfrac{9}{10}\)
`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)
`\Rightarrow x=`\(\dfrac{11}{10}\)
Vậy, `x=`\(\dfrac{11}{10}\)
