(\(\frac{-2x}{5}\)+1)\(\div\)(-5)\(=\)\(\frac{-1}{25}\)
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14 tháng 10 2016
Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
15 tháng 10 2016
\(E=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(E=\frac{\frac{4}{5}:\frac{4}{5}:1,25}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{36}{17}}+\frac{6}{5}.\frac{1}{2}.\frac{5}{4}\)
\(E=\frac{1:\frac{5}{4}}{\frac{3}{5}}+\frac{1.\frac{7}{4}}{\frac{119}{36}.\frac{36}{17}}+\frac{3}{4}\)
\(E=\frac{4}{5}.\frac{5}{3}+\frac{\frac{7}{4}}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{7}{4}.\frac{1}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
\(E=\frac{4}{3}+1=\frac{7}{3}\)
NH
1
PT
16 tháng 1 2017
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+...+\frac{1}{243}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+...+\frac{1}{243}\right)\)
\(\Rightarrow A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81\)
\(=\frac{1}{3}.5\)
\(=\frac{5}{3}\)
\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)
\(\Rightarrow A>\frac{5}{4}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{5}{4}\)
\(\Rightarrow1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{9}{4}\)
\(\Rightarrow\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}\right)>\frac{9}{4}.\frac{1}{5}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\)
XG
2
1 tháng 6 2020
\(=\frac{4}{3}.\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right)\div\frac{3}{5}\)
\(=1-\frac{4}{5}\div\frac{3}{5}\)
\(=1-\frac{4}{5}.\frac{5}{3}\)
\(=1-\frac{4}{3}=-\frac{1}{3}\)
\(\left(\frac{-2x}{5}+1\right):\left(-5\right)=-\frac{1}{25}\)
\(\Leftrightarrow\frac{-2x+5}{5}=-\frac{1}{25}\cdot\left(-5\right)=\frac{1}{5}\)
\(\Leftrightarrow-2x+5=1\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)