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a) x.(\(\dfrac{6}{7}\)+\(\dfrac{5}{6}\))=\(\dfrac{3}{4}\)
x.\(\dfrac{71}{42}\)=\(\dfrac{3}{4}\)
x=\(\dfrac{3}{4}\):\(\dfrac{71}{42}\)
x=\(\dfrac{63}{142}\)
a.\(\dfrac{6}{7}x+\dfrac{5}{6}x=\dfrac{3}{4}\)
\(x.\left(\dfrac{6}{7}+\dfrac{5}{6}\right)=\dfrac{3}{4}\)
\(x.\dfrac{71}{42}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{71}{42}\)
\(x=\dfrac{63}{142}\)
b\(\dfrac{5}{4}-\dfrac{3}{5}:x=1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{5}{4}-1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{-1}{12}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{12}\)
\(x=\dfrac{-36}{5}\)
c. \(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right):3\dfrac{1}{2}=0,5\)
\(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right)=0,5:3\dfrac{1}{2}\)
\(\dfrac{4}{7}x-\dfrac{1}{3}=\dfrac{1}{7}\)
\(\dfrac{4}{7}x=\dfrac{1}{7}+\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{10}{21}\)
\(x=\dfrac{10}{21}:\dfrac{4}{7}\)
\(x=\dfrac{5}{6}\)
d.\(\dfrac{4}{5}-\dfrac{2}{3}x=1\dfrac{1}{4}+2,5x\)
\(\dfrac{4}{5}-\left(\dfrac{2}{3}x-2,5x\right)=1\dfrac{1}{4}\)
\(\dfrac{4}{5}-\dfrac{-11}{6}x=1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{4}{5}-1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{-9}{20}\)
\(x=\dfrac{-9}{20}:\dfrac{-11}{6}\)
\(x=\dfrac{27}{110}\)
có sai sót j xin bn thông cảm !
tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
A,
\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)
\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)
B,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
a.\(\dfrac{-4}{5}-\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{2}{7}\)
\(\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{-4}{5}-\dfrac{2}{7}=\dfrac{-38}{35}\)
\(\dfrac{2}{3}x=\dfrac{-38}{35}-1\dfrac{1}{4}\)
\(\dfrac{2}{3}x=\dfrac{-327}{140}\Rightarrow x=\dfrac{-327}{140}:\dfrac{2}{3}=\dfrac{-981}{280}\)
Vậy \(x=\dfrac{-981}{280}\)
b. \(\dfrac{5}{6}+\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}-\dfrac{5}{6}=\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{3}{4}-\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{9}{4}\Rightarrow x=\dfrac{1}{2}:\dfrac{9}{4}=\dfrac{2}{9}\)
Vậy \(x=\dfrac{2}{9}\)
c. \(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right):\dfrac{3}{4}=0,7\)
\(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right)=0,7.\dfrac{3}{4}=\dfrac{21}{40}\)
\(\dfrac{4}{5}x=\dfrac{21}{40}+1\dfrac{1}{3}=\dfrac{223}{120}\)
\(\Rightarrow x=\dfrac{223}{120}:\dfrac{4}{5}=\dfrac{223}{96}\)
Vậy \(x=\dfrac{223}{96}\)
d. \(\dfrac{5}{6}-\dfrac{3}{4}x=1\dfrac{1}{3}+0,5x\)
\(0,5x+\dfrac{3}{4}x=\dfrac{5}{6}-1\dfrac{1}{3}\)
\(\dfrac{5}{4}x=\dfrac{-1}{2}\Rightarrow x=\dfrac{-1}{2}:\dfrac{5}{4}=\dfrac{-2}{5}\)
Vậy \(x=\dfrac{-2}{5}\)
a: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)
b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)
c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)
\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)
d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)
\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\)
b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\)
\(\dfrac{x}{3}=\dfrac{-14}{15}\)
\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)
\(x=\dfrac{-2}{7}+\dfrac{9}{7}\)
\(x=1\)