A = 1 1/10 + 2 2/10 + ..... + 9 9/10
GIÚP MIK
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Ta có :
\(A=10+10^2+10^3+...+10^{2018}\)
\(10A=10^2+10^3+10^4+...+10^{2019}\)
\(10A-A=\left(10^2+10^3+10^4+...+10^{2019}\right)-\left(10+10^2+10^3+...+10^{2018}\right)\)
\(9A=10^{2019}-10\)
\(A=\frac{10^{2019}-10}{9}\)
Vì \(\frac{10^{2019}-10}{9}>\frac{1}{9}\)\(\Rightarrow\)\(A>\frac{1}{9}\)\(\Rightarrow\)ĐỀ SAI
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{10}{10!}-\frac{1}{10!}\)
\(A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+...+\frac{1}{9!}-\frac{1}{10!}\)
\(A=1-\frac{1}{10!}\)
\(\Rightarrow A< 1\left(đpcm\right)\)
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
a) nhân ra thôi b
\(=\frac{\left(2\sqrt{10}-5\right)\left(9+\sqrt{10}\right)}{71}=\frac{18\sqrt{10}-45+20-5\sqrt{10}}{71}=\frac{-25+13\sqrt{10}}{71}.\)
b)cách khác nhé !\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}.\)
\(2\frac{1}{6}x3+3\frac{4}{5}x10x9\frac{1}{10}\)
\(=1+38x9\frac{1}{10}\)
\(=1+345,8\)
\(346,8\)
\(A=1\frac{1}{10}+2\frac{2}{10}+...+9\frac{9}{10}\)
\(\Leftrightarrow A=\frac{11}{10}+\frac{22}{10}+...+\frac{99}{10}\)
\(\Leftrightarrow A=\frac{11+22+...+88+99}{10}\)
\(\Leftrightarrow A=\frac{\left(99+11\right).9}{20}\)
\(\Leftrightarrow A=\frac{990}{20}\)
\(\Leftrightarrow A=49,5\)
\(1\frac{1}{10}+2\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+...+9\frac{9}{10}\)
\(=\frac{11}{10}+\frac{22}{10}+\frac{33}{10}+\frac{44}{10}+...+\frac{99}{10}\)
\(=1,1+2,2+3,3+4,4+...+9,9\)
\(=\frac{\left(9,9+1,1\right).9}{2}=49.5\)