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Gọi \(S=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)
\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)
Nhân hai vế với 3 và áp dụng công thức tách một phân số thành hiệu hai phân số:
\(\dfrac{x}{n\left(n+x\right)}=\dfrac{1}{n}-\dfrac{1}{n+x}\)
\(\Rightarrow3S=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\)
\(\Rightarrow3S=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\)
\(\Rightarrow3S=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\)
\(\Rightarrow3S=\dfrac{1}{2}-\dfrac{1}{20}\)
\(\Rightarrow3S=\dfrac{10}{20}-\dfrac{1}{20}\)
\(\Rightarrow3S=\dfrac{9}{20}\)
\(\Rightarrow S=\dfrac{9}{20}:3\)
\(\Rightarrow S=\dfrac{9}{20}.\dfrac{1}{3}\)
\(\Rightarrow S=\dfrac{3}{20}\)
Đặt \(A=\dfrac{1}{10}-\dfrac{1}{40}-\dfrac{1}{88}-\dfrac{1}{154}-\dfrac{1}{238}-\dfrac{1}{340}\)
\(\Leftrightarrow3A=\dfrac{3}{10}-\dfrac{3}{40}-\dfrac{3}{88}-\dfrac{3}{154}-\dfrac{3}{238}-\dfrac{3}{340}\)
\(\Leftrightarrow3A=\dfrac{3}{2\cdot5}-\dfrac{3}{5\cdot8}-\dfrac{3}{8\cdot11}-\dfrac{3}{11\cdot14}-\dfrac{3}{14\cdot17}-\dfrac{3}{17\cdot20}\)
\(\Leftrightarrow3A=\left(\dfrac{1}{2}-\dfrac{1}{5}\right)-\left(\dfrac{1}{5}-\dfrac{1}{8}\right)-\left(\dfrac{1}{8}-\dfrac{1}{11}\right)-\left(\dfrac{1}{11}-\dfrac{1}{14}\right)-\left(\dfrac{1}{14}-\dfrac{1}{17}\right)-\left(\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{8}-\dfrac{1}{8}+\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{14}-\dfrac{1}{14}+\dfrac{1}{17}-\dfrac{1}{17}+\dfrac{1}{20}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{20}\\ \Leftrightarrow3A=\dfrac{3}{20}\\ \Leftrightarrow A=\dfrac{1}{20}\)
a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)
b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)
c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
A=1/5^3+1/6^3+...+1/2023^3
1/5^3<1/4*5*6
Xét tương tự, ta đều sẽ được:
\(\dfrac{1}{n^3}< \dfrac{1}{n\left(n-1\right)\left(n+1\right)}\)
=>\(A< \dfrac{1}{4\cdot5\cdot6}+\dfrac{1}{5\cdot6\cdot7}+...+\dfrac{1}{2022\cdot2023\cdot2024}\)
=>\(A< \dfrac{1}{2}\left(\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+...+\dfrac{2}{2022\cdot2023\cdot2024}\right)\)
=>\(A< \dfrac{1}{2}\left(\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}+...+\dfrac{1}{2022\cdot2023}-\dfrac{1}{2023\cdot2024}\right)\)
=>A<1/40
Ta có BĐT: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}< \dfrac{1}{k^3}< \dfrac{1}{\left(k-1\right)\cdot k\cdot\left(k+1\right)}\)
Do đó, ta được:
\(\dfrac{1}{5\cdot6\cdot7}+\dfrac{1}{6\cdot7\cdot8}+...+\dfrac{1}{2023\cdot2024\cdot2025}< A\)
\(\Leftrightarrow A>\dfrac{1}{2}\left(\dfrac{1}{5\cdot6}-\dfrac{1}{2024\cdot2025}\right)>\dfrac{1}{2}\left(\dfrac{1}{30}-\dfrac{1}{390}\right)=\dfrac{1}{65}\)
=>1/65<A<1/40
Bài 1:
\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)
\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\)
Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\)
\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\)
\(\dfrac{-18}{31}\)
\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\)
Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)
Bài 2:
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
\(\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}\)+\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{3}\right)}\)
=\(\frac{3}{5}\)+\(\frac{1}{\frac{5}{2}}\)
=\(\frac{3}{5}\)+\(\frac{2}{5}\)
=1 !!!
ồ, lâu h ms gặp
a,
Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)
Ta có:
\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
b,
\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)
Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)
c,
\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)
Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)
Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)
\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)
\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)
\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)
\(B< A\)
\(C=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)
\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)
\(C=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\right)\)
\(C=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(C=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(C=\dfrac{1}{3}.\dfrac{9}{20}\)
\(C=\dfrac{3}{20}\)