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\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=2008\cdot\frac{47}{3}\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=\frac{94376}{3}\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=2009-\frac{94376}{3}\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=-\frac{88349}{3}\)
\(\frac{41}{9}+x\) \(=-\frac{88349}{3}+\frac{133}{18}\)
\(\frac{41}{9}+x\) \(=-\frac{529961}{18}\)
\(x=-\frac{529961}{18}-\frac{41}{9}\)
\(x=-\frac{176681}{6}\)
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
a, \(x-\frac{1}{3}=\frac{3}{4}\)
\(x=\frac{3}{4}+\frac{1}{3}\)
\(x=\frac{9}{13}+\frac{4}{12}=\frac{13}{12}\)
a. \(x=\frac{3}{4}+\frac{1}{3}\)
\(x=\frac{10}{12}\)
b. \(x+\frac{5}{2}=\frac{15}{4}\)
\(x=\frac{15}{4}-\frac{5}{2}\)
\(x=\frac{5}{4}\)
c. \(x:\frac{1}{4}=\frac{37}{2}\)
\(x=\frac{37}{2}\)x \(\frac{1}{4}\)
\(x=\frac{37}{8}\)
d. \(x=\frac{2}{9}:\frac{2}{3}\)
\(x=\frac{2}{9}\)x \(\frac{3}{2}\)
\(x=\frac{1}{3}\)
=> 49/9 + x = ( 1. 63/4) + 133/18 = 833/36
=> x = 833/36 - 49/9 = 637/36