\(\frac{5+...}{11+...}=\frac{1}{2}\)
\(\frac{5-...}{9-...}=\frac{1}{2}\)
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\(\frac{16}{5}.\frac{5}{11}+\frac{17}{5}.\frac{5}{11}-\frac{1}{2}\)
\(=\frac{5}{11.}\left(\frac{16}{5}+\frac{17}{5}\right)-\frac{1}{2}\)
\(=\frac{5}{11}.\frac{33}{5}-\frac{1}{2}\)
\(=3-\frac{1}{2}\)
\(=\frac{5}{2}\)
~Hok tốt~
\(\frac{3}{4}\)-\(\frac{-5}{9}\)-\(\frac{11}{36}\)=\(\frac{27}{36}\)-\(\frac{-20}{36}\)-\(\frac{11}{36}\)=1
\(\frac{1}{9}\)+\(\frac{-5}{3}\)-\(\frac{-13}{18}\)=\(\frac{2}{18}\)+\(\frac{-30}{18}\)-\(\frac{-13}{18}\)=\(\frac{-15}{18}\)=\(\frac{-5}{6}\)
\(\frac{3}{4}-\frac{-5}{9}-\frac{11}{36}=\frac{27}{36}-\frac{-20}{36}-\frac{11}{36}=\frac{47}{36}-\frac{11}{36}=\frac{36}{36}=1\)
\(\frac{1}{9}+\frac{-5}{3}-\frac{13}{18}=\frac{2}{18}+\frac{-30}{18}-\frac{13}{18}=\frac{-28}{18}+\frac{13}{18}=\frac{-15}{18}=\frac{-5}{6}\)
\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
= \(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
= \(\left(\frac{3}{4}+\frac{11}{4}+\frac{1}{2}\right)\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3\)
= \(8+2+1-3\)
= \(8\)
#)Giải :
\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
\(=\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
\(=\left(\frac{3}{4}+\frac{11}{4}\right)+\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3+\frac{1}{2}\)
\(=\frac{7}{2}+2+1-3+\frac{1}{2}\)
\(=\frac{7}{2}+\frac{1}{2}\)
\(=4\)
\(\Leftrightarrow\left(\frac{3}{4}x-\frac{9}{16}\right)\left(\frac{1}{3}-\frac{3}{5}.\frac{1}{x}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{4};\frac{9}{5}\right\}\)
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B