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Ta có : \(\dfrac{1}{1794}\)>\(\dfrac{1}{1795^2}\)
\(\dfrac{1}{1794}\)>\(\dfrac{1}{1796^2}\)
\(\dfrac{1}{1794}\)>\(\dfrac{1}{1797^2}\)
.....................
\(\dfrac{1}{1794}\)>\(\dfrac{1}{2016^2}\)
\(\dfrac{1}{1794}\)>\(\dfrac{1}{2017^2}\)
\(\Leftrightarrow\)\(\dfrac{1}{1794}\)>\(\dfrac{1}{1795^2}\)+\(\dfrac{1}{1796^2}\)+\(\dfrac{1}{1797^2}\)+. . .+\(\dfrac{1}{2016^2}\)+\(\dfrac{1}{2017^2}\)
\(\frac{1}{1975^2}+\frac{1}{1976^2}+...+\frac{1}{2017^2}< \frac{1}{1974.1975}+\frac{1}{1975.1976}+...+\frac{1}{2016.2017}\)
\(=\frac{1}{1974}-\frac{1}{1975}+\frac{1}{1975}-\frac{1}{1976}+...+\frac{1}{2016}-\frac{1}{2017}=\frac{1}{1974}-\frac{1}{2017}< \frac{1}{1974}\)
a, sai đề
b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )
\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Câu a thiếu đề rồi bạn ơi mik giải câu b đây:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)
\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Bài 1: Ta có:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
Vậy \(M=\dfrac{5}{16}\)
Bài 2: Ta có:
\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)
\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)
\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)
Vậy \(A=\dfrac{1}{10}\)
Giải:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)
\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)
\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)
\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)
\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)
\(M=\dfrac{120-\dfrac{1}{2}.40.5.\dfrac{1}{5}.20.\dfrac{1}{4}-20}{1+5+....+41}\\ =\dfrac{120-20.5-20}{1+5+...+41}\\ =\dfrac{0}{1+5+...+41}\\ =0\)
\(N=10101\left(\dfrac{6}{3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}-\dfrac{7}{3.7.11.13.37}\right)\\ =10101\left(-\dfrac{2}{2.3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}\right)\\ =3.7.13.37\left(\dfrac{4}{2.3.7.11.13.37}\right)=\dfrac{4}{2.11}=\dfrac{2}{11}\)
Ta có:
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
Vì \(\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
Ta có \(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(\Leftrightarrow A=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
\(\Leftrightarrow B=1+\dfrac{2}{20^{10}-3}\)
Vì 1=1 mà\(20^{10}-1>20^{10}-3\Rightarrow\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\Rightarrow1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\)
hay A < B
Vậy A < B
\(20M=\dfrac{20^{1976}+1+19}{20^{1976}+1}=1+\dfrac{19}{20^{1976}+1}\)
\(20N=\dfrac{20^{1977}+1+19}{20^{1977}+1}=1+\dfrac{19}{20^{1977}+1}\)
mà \(20^{1976}+1< 20^{1977}+1\)
nên M>N