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2.
\(A=\dfrac{36}{1\cdot3\cdot5}+\dfrac{36}{3\cdot5\cdot7}+...+\dfrac{36}{25\cdot27\cdot29}\\ =9\cdot\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\\ =9\cdot\dfrac{1}{3}-9\cdot\dfrac{1}{783}\\ =3-\dfrac{1}{87}< 3\)
Vậy \(A< 3\)
b,
\(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{50}\\ B< 2-\dfrac{1}{50}< 2\)
Vậy \(B< 2\)
\(P=\dfrac{2}{60\cdot63}+\dfrac{2}{63\cdot66}+...+\dfrac{2}{117\cdot120}+\dfrac{2}{2011}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)\)
\(Q=\dfrac{5}{40\cdot44}+\dfrac{5}{44\cdot48}+...+\dfrac{5}{76\cdot80}+\dfrac{5}{2011}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\)
\(\dfrac{3}{2011}< \dfrac{4}{2011}\Rightarrow\dfrac{1}{2}+\dfrac{3}{2011}< \dfrac{1}{2}+\dfrac{4}{2011}\left(1\right)\)
\(\dfrac{2}{3}< \dfrac{5}{4}\left(2\right)\)
Từ (1) và (2) ta có: \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)< \dfrac{5}{4}\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\Leftrightarrow P< Q\)
Vậy P < Q
\(C=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)
\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)
\(C=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\right)\)
\(C=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(C=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(C=\dfrac{1}{3}.\dfrac{9}{20}\)
\(C=\dfrac{3}{20}\)
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) \(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{12}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{4}+\dfrac{-2}{5}=\dfrac{-3}{20}\)
b) \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}=\left(\dfrac{-2}{3}-\dfrac{5}{6}\right)+\left(\dfrac{-1}{5}-\dfrac{-7}{10}\right)+\dfrac{3}{4}\)
\(=\dfrac{-3}{2}+\dfrac{1}{2}-\dfrac{3}{4}\)
= \(=-1-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
c)\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)
= \(\left(\dfrac{1}{2}-\dfrac{-1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-4}{35}+\dfrac{5}{7}-\dfrac{-2}{5}\right)+\dfrac{1}{41}\)
= \(1+1+\dfrac{1}{41}\)
= \(\dfrac{83}{41}\)
d)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
= \(\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{1}\)
= \(\dfrac{1}{100}-\dfrac{1}{1}\)
= \(\dfrac{-99}{100}\)
d đảo 1/1.2.1/2.3 ... 1/99.1000
=1/1 -1/2 +1/2-1/3 ... -1/99 - 1/1000
=1/1 -1/1000
=999/1000
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
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: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)