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\(\frac{18}{11}+\left(\frac{7}{4}-\frac{3}{5}\right):\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}:\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}\times\frac{2}{1}\)
\(\frac{18}{11}+\frac{23}{10}\)
\(\frac{433}{110}\)
\(3-x\times\frac{1}{2}=\frac{3}{4}\)
\(x\times\frac{1}{2}=3-\frac{3}{4}\)
\(x\times\frac{1}{2}=\frac{12}{4}-\frac{3}{4}\)
\(x\times\frac{1}{2}=\frac{9}{4}\)
\(x=\frac{9}{4}\div\frac{1}{2}\)
\(x=\frac{9}{4}\times\frac{2}{1}\)
\(x=\frac{18}{4}=\frac{9}{2}\)
Vậy x = 9/2
ik r mk làm tiếp cho
a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
\(\left(a\right)37064-64\left(82+\frac{42966}{217}\right)\)
\(=37064-64\left(82+198\right)\)
\(=37064-64\cdot280\)
\(=30764-17920=19144\)
\(\left(b\right)320-\left(120,5+95,25+5,25\right)+\frac{84}{12}\cdot12,5\)
\(=320-\left(120,5+100,5\right)+7\cdot12,5\)
\(=320-221+87,5=11,5\)
\(\left(c\right)\left(4\frac{2}{5}+2\frac{3}{7}\right)-\left(2\frac{2}{5}-5\frac{4}{7}\right)\)
\(=4\frac{2}{5}-2\frac{2}{5}+2\frac{3}{7}+5\frac{4}{7}\)
\(=2+8=10\)
\(\left(d\right)\left(\frac{1998}{18}-\frac{1443}{13}\right)\left(16996-\frac{1110}{30}\cdot305\right)\)
\(=\left(111-111\right)\left(16996-\frac{1110}{30}\cdot305\right)\)
\(=0\left(16996-\frac{1110}{30}\cdot305\right)=0\)
\(\left(e\right)5\frac{3}{5}+1,75+6\frac{1}{8}+4\frac{1}{4}+3,875+3,4\)
\(=5\frac{3}{5}+1\frac{3}{4}+6\frac{1}{8}+4\frac{1}{4}+3\frac{7}{8}+3\frac{2}{5}\)
\(=5\frac{3}{5}+3\frac{2}{5}+1\frac{3}{4}+4\frac{1}{4}+6\frac{1}{8}+3\frac{7}{8}\)
\(=9+6+10=25\)