23/x + 3/4 = -1/2 : 5
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
A=((2-2/19+2/23-1/1010)/(3-3/19+3/23-3/2020))x((4-4/29+4/41-1/505)/(5-5/29+5/41-1/404))
giúp mình với
\(\dfrac{-9}{46}-4\dfrac{1}{23}:\left(3\dfrac{1}{4}-x:\dfrac{3}{5}\right)+2\dfrac{8}{23}=1\)
\(\dfrac{-9}{46}+2\dfrac{8}{23}-4\dfrac{1}{23}:\left(3\dfrac{1}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{-9}{46}+\dfrac{54}{23}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{-9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{99}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{99}{46}-1\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{99}{46}-\dfrac{46}{46}\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{53}{46}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{93}{23}:\dfrac{53}{46}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{93}{23}\cdot\dfrac{46}{53}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{186}{53}\)
\(x:\dfrac{3}{5}=\dfrac{13}{4}-\dfrac{186}{53}\)
\(x:\dfrac{3}{5}=\dfrac{689}{212}-\dfrac{744}{212}\)
\(x:\dfrac{3}{5}=\dfrac{-55}{212}\)
\(x=\dfrac{-55}{212}\cdot\dfrac{3}{5}\)
\(x=\dfrac{-33}{212}\)
Vậy \(x=\dfrac{-33}{212}\).