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#muon roi ma sao con
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\right)=0\Leftrightarrow x=-100\)
Vậy x = -100
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
\(\frac{1}{1}\)x 2 x 3 + \(\frac{1}{2}\)x 3 x 4 + \(\frac{1}{3}\)x 4 x 5 + \(\frac{1}{4}\)x 5 x 6
= 1 x 2 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)x 6
= 2 +\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ 1, 5
=
\(\frac{3}{7}:x+\frac{1}{4}=\frac{5}{9}\)
\(\frac{3}{7}:x=\frac{5}{9}-\frac{1}{4}=\frac{20-9}{36}=\frac{11}{36}\)
\(x=\frac{3}{7}:\frac{11}{36}=\frac{3}{7}\times\frac{36}{11}=\frac{3\times36}{7\times11}=\frac{108}{77}\)
Vậy \(x=\frac{108}{77}\)
\(\frac{3}{7}:x+\frac{1}{4}=\frac{5}{9}\)
\(\frac{3}{7}:x=\frac{5}{9}-\frac{1}{4}\)
\(\frac{3}{7}:x=\frac{11}{36}\)
\(x=\frac{3}{7}:\frac{11}{36}\)
\(x=\frac{108}{77}\)