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. Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
.................
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{11}+\dfrac{1}{12}+......+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+.....+\dfrac{1}{20}\)
\(\Leftrightarrow S>\dfrac{1}{20}.10\)
\(\Leftrightarrow S>\dfrac{1}{2}\)
2. \(\dfrac{x}{12}=\dfrac{-1}{24}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x}{12}=-\dfrac{1}{6}\)
\(\Leftrightarrow6x=-12\)
\(\Leftrightarrow x=-2\)
Vậy ...
3. \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+........+\dfrac{2}{19.21}\)
\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{19}-\dfrac{1}{21}\)
\(=\dfrac{1}{5}-\dfrac{1}{21}\)
\(=\dfrac{16}{105}\)
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
= \(\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+1-\dfrac{1}{4}+...+1-\dfrac{1}{100}}\)
= \(\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
=\(\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)= 2
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{2017}}\)
Vậy \(A=1-\dfrac{1}{2^{2017}}\)
b) \(1-\dfrac{1}{2^{2017}}< 1\Rightarrow A< 1\)
Vậy A < 1
Ta có : A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{9}{9}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\) (1)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{5}{10}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{4}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\) (2)
Từ (1) và (2)\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
\(x:4\dfrac{1}{3}=2,5\\ x:\dfrac{13}{3}=\dfrac{5}{2}\\ x=\dfrac{5}{2}.\dfrac{13}{3}\\ x=\dfrac{65}{6}=10\dfrac{5}{6}\)
\(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\cdot\cdot\cdot\left(1+\dfrac{7}{180}\right)=\dfrac{16}{9}\cdot\dfrac{27}{20}\cdot\cdot\cdot\dfrac{187}{180}=\dfrac{2.8}{1\cdot9}\cdot\dfrac{3\cdot9}{2\cdot10}\cdot\cdot\cdot\dfrac{11\cdot17}{10\cdot18}=\dfrac{\left(2\cdot3\cdot...\cdot11\right)\cdot\left(8\cdot9\cdot...\cdot17\right)}{\left(1\cdot2\cdot...\cdot10\right)\cdot\left(9\cdot10\cdot...\cdot18\right)}=\dfrac{11\cdot8}{1\cdot18}=\dfrac{88}{18}=\dfrac{44}{9}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
Cộng vế với vế ta được
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow T< 2-\dfrac{1}{20}=\dfrac{39}{20}\)
mà 39/20 < 8/7 => T < 8/7