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c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)
vậy x = \(\frac{262}{35}\)
d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\)
vậy x = \(\frac{57}{8}\)
e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)
vậy x = \(\frac{1}{4}\)
a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)
vậy x = \(\frac{35}{12}\)
b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)
vậy x = \(\frac{19}{14}\)
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
\(x-\frac{3}{4}-x.\frac{2}{3}+x:\frac{1}{2}-x:\frac{2}{5}=\frac{11}{4}\)
\(x-x.\frac{2}{3}+x.2-x.\frac{5}{2}=\frac{11}{4}+\frac{3}{4}\)
\(x\left(1-\frac{2}{3}+2-\frac{5}{2}\right)=\frac{7}{2}\)
\(x.\frac{-1}{6}=\frac{7}{2}\)
\(x=\frac{7}{2}:-\frac{1}{6}\)
\(x=-21\)
Vậy \(x=-21\)
\((2,7.x-1\frac{1}{2})\div\frac{2}{7}=\frac{-21}{4}\) \(3\frac{1}{3}.x+16\frac{3}{4}=-13.25\)
\(2,7.x-1\frac{1}{2}=-\frac{21}{4}\cdot\frac{2}{7}\) \(\frac{10}{3}.x+\frac{67}{4}=-13.25\)
\(2,7.x-\frac{3}{2}=-\frac{3}{2}\) \(\frac{10}{3}.x+\frac{67}{4}=-\frac{53}{4}\)
\(2,7.x=-\frac{3}{2}+\frac{3}{2}\) \(\frac{10}{3}.x=-\frac{53}{4}-\frac{67}{4}\)
\(2,7.x=0\) \(\frac{10}{3}.x=-30\)
\(x=0:2,7\) \(x=-30:\frac{10}{3}\)
\(x=0\) \(x=-9\)
Vậy x=0 Vậy x= -9
\(\left(4.5-2.x\right):\frac{3}{4}=1\frac{1}{3}\) \(1.5+1\frac{1}{4}.x=\frac{2}{3}\)
\(\left(4.5-2.x\right)=1\frac{1}{3}\cdot\frac{3}{4}\) \(1\frac{1}{4}.x=\frac{2}{3}-1.5\)
\(4.5-2.x=\frac{4}{3}\cdot\frac{3}{4}\) \(\frac{5}{4}.x=\frac{2}{3}-\frac{3}{2}\)
\(4.5-2.x=1\) \(\frac{5}{4}.x=-\frac{5}{6}\)
\(2.x=4.5-1\) \(x=-\frac{5}{6}:\frac{5}{4}\)
\(2.x=3.5\) \(x=-\frac{2}{3}\)
\(x=3.5:2\)
\(x=1.75\) Vậy \(x=-\frac{2}{3}\)
Vậy x=1.75
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
a)
\(x+\frac{11}{4}=\frac{17}{3}\)
\(x=\frac{17}{3}-\frac{11}{4}\)
\(x=\frac{35}{12}\)
b)
\(x-\frac{9}{5}=\frac{23}{7}\)
\(x=\frac{23}{7}+\frac{9}{5}\)
\(x=\frac{178}{35}\)
c)
\(x\)x \(\frac{7}{2}=\frac{19}{4}\)
\(x=\frac{19}{4}:\frac{7}{2}\)
\(x=\frac{19}{14}\)
d)
\(x:\frac{8}{3}=\frac{13}{3}\)
\(x=\frac{13}{3}\)x \(\frac{8}{3}\)
\(x=\frac{104}{9}\)
\(\frac{3}{4}.\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{3}{5}-x=\frac{2}{3}\)
\(x=\frac{3}{5}-\frac{2}{3}\)
\(x=-\frac{1}{15}\)
Vậy \(x=-\frac{1}{15}\)
\(x+\frac{1}{2}.\frac{2}{3}=\frac{3}{4}\)
\(x+\frac{1}{3}=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{1}{3}\)
\(x=\frac{5}{12}\)
vậy \(x=\frac{5}{12}\)
hơ hơ =v
\(\frac{3}{4}\times\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{3}{5}-x=\frac{2}{3}\)
\(x=\frac{3}{5}-\frac{2}{3}\)
\(x=\frac{-1}{15}\)
a) \(\frac{a}{b}x-\frac{7}{8}=\frac{1}{4}\)
\(\Rightarrow\frac{a}{b}x=\frac{1}{4}+\frac{7}{8}\)
\(\Rightarrow\frac{a}{b}x=\frac{9}{8}\)
\(\Rightarrow x=\frac{9}{8}:\frac{a}{b}=\frac{9}{8}.\frac{b}{a}\)
\(\Rightarrow x=\frac{9b}{8a}\)
b) \(\frac{3}{2}x-\frac{1}{2}=\frac{1}{3}:\left(\frac{-5}{6}\right)\)
\(\Rightarrow\frac{3}{2}x-\frac{1}{2}=\frac{-2}{5}\)
\(\Rightarrow\frac{3}{2}x=\frac{-2}{5}+\frac{1}{2}\)
\(\Rightarrow\frac{3}{2}x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\frac{3}{2}\)
\(\Rightarrow x=\frac{1}{15}\)
c) \(\frac{2}{3}\left(x+\frac{5}{4}\right)-\frac{1}{3}\left(\frac{2}{3}-x\right)=\frac{4}{3}\)
\(\Rightarrow\frac{2}{3}x+\frac{5}{6}-\frac{2}{9}+\frac{1}{3}x=\frac{4}{3}\)
\(\Rightarrow\frac{2}{3}x+\frac{1}{3}x=\frac{4}{3}-\frac{5}{6}+\frac{2}{9}\)
\(\Rightarrow x=\frac{13}{18}\)
a) \(x=\frac{7}{20}\)
b) \(x=\frac{7}{12}\)
c)\(x=\frac{8}{15}\)
a ) \(\frac{7}{8}:x=3-\frac{1}{2}\)
\(\frac{7}{8}:x=\frac{5}{2}\)
\(x=\frac{7}{8}:\frac{5}{2}\)
\(x=0,35\)
b ) \(x+\frac{1}{2}.\frac{1}{3}=\frac{3}{4}\)
\(x+\frac{1}{6}=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{1}{6}\)
\(x=\frac{7}{12}\)
c ) \(\frac{3}{2}.\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{7}{10}-x=\frac{2}{3}\)
\(x=\frac{7}{10}-\frac{2}{3}\)
\(x=\frac{1}{30}\)
d ) \(x.3\frac{1}{3}=3\frac{1}{3}:4\frac{1}{4}\)
\(x:\frac{10}{3}=\frac{10}{3}:\frac{17}{4}\)
\(x:\frac{10}{3}=\frac{40}{51}\)
\(x=\frac{40}{51}:\frac{10}{3}\)
\(x=\frac{4}{17}\)
e ) \(5\frac{2}{3}:x=3\frac{2}{3}-2\frac{1}{2}\)
\(\frac{17}{3}:x=\frac{11}{3}-\frac{5}{2}\)
\(\frac{17}{3}:x=\frac{7}{6}\)
\(x=\frac{17}{3}:\frac{7}{6}\)
\(x=\frac{34}{7}\)
Nếu mình đúng thì các bạn k mình nhé
\(\frac{x+1}{3}=\frac{x-2}{4}\)
=> \(4\left(x+1\right)=3\left(x-2\right)\)
=> \(4x+4=3x-6\)
=> \(4x-3x=-6-4\)
=> \(x=-10\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x+1}{3}=\frac{x-2}{4}=\frac{x+1-\left(x-2\right)}{3-4}=\frac{x+1-x+2}{-1}=\frac{3}{-1}=-3\)
\(\Rightarrow\hept{\begin{cases}x+1=-3.3=-9\\x-2=-3.4=-12\end{cases}}\Leftrightarrow x=-10\)