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a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004
1 x 2 x 3 x 4 x .............x 2002 x 2003
2 x 3 x 4 x .............x 2003 x 2004
1
2004
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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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 \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)
a, \(\Leftrightarrow2x^2=72\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm6\)
Vậy ...
\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)
\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)
Vậy ...
\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)
Vậy ...
\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)
\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)
Vậy ...
a) 2x2 - 72 = 0
\(\Rightarrow\) 2x2 = 72
\(\Rightarrow\) x2 = 36 = 62 = (- 6)2
\(\Rightarrow\) x = 6 hoặc x = - 6
Vậy x = 6 hoặc x = - 6
b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)
\(\Rightarrow\) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)
\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)
\(\Rightarrow\) x = \(\dfrac{13}{4}\)
Vậy x = \(\dfrac{13}{4}\)
`@` `\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=`\(\dfrac{11}{10}\)