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a.\(\dfrac{5}{3}x-2\dfrac{1}{3}=-\dfrac{4}{3}.1\dfrac{1}{8}-\dfrac{2}{3}\)
\(\dfrac{5}{3}x-2\dfrac{1}{3}=\dfrac{5}{6}\)
\(\dfrac{5}{3}x=\dfrac{5}{6}+2\dfrac{1}{3}\)
\(\dfrac{5}{3}x=\dfrac{19}{6}\)
\(x=\dfrac{19}{5}:\dfrac{5}{3}\)
\(x=\dfrac{19}{10}\)
b. \(2\dfrac{1}{6}:x-\dfrac{-5}{8}=\dfrac{-7}{15}:4\dfrac{1}{5}-\dfrac{-6}{7}\)
\(2\dfrac{1}{6}:x+\dfrac{5}{8}=\dfrac{47}{63}\)
\(2\dfrac{1}{6}:x=\dfrac{47}{63}-\dfrac{5}{8}\)
\(2\dfrac{1}{6}:x=\dfrac{61}{504}\)
\(x=2\dfrac{1}{6}:\dfrac{61}{504}\)
\(x=\dfrac{1092}{61}\)
a, \(\dfrac{5}{3}x-2\dfrac{1}{3}=-\dfrac{4}{3}.1\dfrac{1}{8}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{3}x=-\dfrac{3}{2}-\dfrac{2}{3}+2\dfrac{1}{3}=-\dfrac{3}{2}+2=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{5}{3}=\dfrac{3}{10}\)
b, \(2\dfrac{1}{6}:x-\dfrac{-5}{8}=\dfrac{-7}{15}:4\dfrac{1}{5}-\dfrac{-6}{7}\)
\(\Rightarrow\dfrac{13}{6}:x=-\dfrac{7}{15}:\dfrac{21}{5}+\dfrac{6}{7}-\dfrac{5}{8}\)
\(\Rightarrow\dfrac{13}{6}:x=\dfrac{61}{504}\Rightarrow x=\dfrac{1092}{61}\)
c, \(\left(\dfrac{5}{6}x-0,3\right):2\dfrac{1}{3}=25\%\)
\(\Rightarrow\dfrac{5}{6}x-0,3=\dfrac{1}{4}.\dfrac{7}{3}\)
\(\Rightarrow\dfrac{5}{6}x=\dfrac{7}{12}+0,3=\dfrac{53}{60}\)
\(\Rightarrow x=\dfrac{53}{60}:\dfrac{5}{6}=1,06\)
d, \(\dfrac{4}{7}-\dfrac{2}{3}x=1,5+\dfrac{4}{5}x\)
\(\Rightarrow\dfrac{4}{5}x+\dfrac{2}{3}x=\dfrac{4}{7}-1,5\)
\(\Rightarrow\dfrac{22}{15}x=-\dfrac{13}{14}\Rightarrow x=-\dfrac{195}{308}\)
Chúc bạn học tốt!!!
\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)
=> \(x=56\)
2) => 18x = 18
=> x = 1
3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)
=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)
=> \(x=\dfrac{-1}{2}\)
4) 45%.x =\(\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)
=> \(x=\dfrac{4}{3}\)
bài này áp dụng quy tắc nhân chéo nha :vv
a) \(\dfrac{x}{5}=\dfrac{2}{5}\Leftrightarrow5x=2.5=10\Leftrightarrow x=\dfrac{10}{5}=2\)
b) \(\dfrac{3}{8}=\dfrac{6}{x}\Leftrightarrow3x=6.8=48\Leftrightarrow x=\dfrac{48}{3}=16\)
c)\(\dfrac{1}{9}=\dfrac{x}{27}\Leftrightarrow9x=27\Leftrightarrow x=\dfrac{27}{9}=3\)
d)\(\dfrac{4}{x}=\dfrac{8}{6}\Leftrightarrow8x=4.6=24\Leftrightarrow x=\dfrac{24}{8}=3\)
e) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+4x=-6+20\\ \Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\)
g) \(\dfrac{x}{-2}=\dfrac{-8}{x}\Leftrightarrow x^2=\left(-2\right)\left(-8\right)=16\\ \Rightarrow x=\pm4\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b:
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)
\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)
\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)
a, \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{3}{5}+0,415\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{203}{200}\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{203}{200}+8\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{1803}{200}\)
\(\left(\dfrac{13}{4}-30,75\right).x=\dfrac{1803}{200}\)
\(\dfrac{-55}{2}.x=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}\)
\(x=\dfrac{-1803}{5500}\)
Nếu là tìm số nguyên thì hình như đề sai rồi bạn
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b, \(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
Cho \(A=4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(A=4\dfrac{1}{3}.\dfrac{-1}{3}\)
\(A=\dfrac{13}{3}.\dfrac{-1}{3}\)
\(A=\dfrac{-13}{9}\)
Cho \(B=\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(B=\dfrac{2}{3}.\left(\dfrac{-1}{6}-\dfrac{3}{4}\right)\)
\(B=\dfrac{2}{3}.\dfrac{-11}{12}\)
\(B=\dfrac{-11}{18}\)
Ta có: \(A\le x\le B\)
\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
a)\(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)
\(\dfrac{2}{3}x=\dfrac{25}{12}\)
\(x=\dfrac{25}{12}:\dfrac{2}{3}\)
=>\(x=\dfrac{25}{8}\)
a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)
\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)
\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)
\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)
d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=2\)
\(\Rightarrow x=2:\dfrac{1}{4}\)
\(\Rightarrow x=2.4=8\)
a/ \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{-5}{12}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{20}\)
\(\Leftrightarrow x=\dfrac{29}{10}\)
Vậy ...
b/ \(\left(4x-3\right)\left(\dfrac{5}{4}x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{8}{5}\end{matrix}\right.\)
Vậy .....
c/ \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)
\(\Leftrightarrow\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=-\dfrac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=-\dfrac{19}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{38}{21}\end{matrix}\right.\)
Vậy ......
d/ \(\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\)
\(\Leftrightarrow\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(\Leftrightarrow\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{9}{10}\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
a. \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)
\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{6}-\dfrac{-5}{12}\)
\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{4}\)
\(\dfrac{3}{6}x=\dfrac{5}{4}+\dfrac{1}{5}\)
\(\dfrac{3}{6}x=\dfrac{29}{20}\)
\(x=\dfrac{29}{20}:\dfrac{3}{6}\)
\(x=\dfrac{29}{10}\)
Vậy...
b. \(\left(4x-3\right).\left(\dfrac{5}{4}x+2\right)=0\)
\(\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-8}{5}\end{matrix}\right.\)
Vậy ...
c. \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)
\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=1,5+\dfrac{3}{4}\)
\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{-9}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=\dfrac{-19}{12}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=\dfrac{-38}{21}\end{matrix}\right.\)
Vậy...
`@` `\text {Ans}`
`\downarrow`
\(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
`=>`\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
`=>`\(x=-\dfrac{13}{24}\)
Vậy, `x =`\(-\dfrac{13}{24}\)
\(x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{13}{24}\)