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\(a.\)
\(\dfrac{3}{16}:\dfrac{?}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{16}\cdot\dfrac{8}{?}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{2?}=\dfrac{3}{4}\)
\(\Leftrightarrow?=2\)
\(b.\)
\(\dfrac{1}{25}:-\dfrac{3}{?}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{1}{25}\cdot\dfrac{-?}{3}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{-?}{75}=-\dfrac{1}{15}\)
\(\Leftrightarrow?=\dfrac{75}{15}=5\)
\(c.\)
\(\dfrac{?}{12}:-\dfrac{4}{9}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{?}{12}\cdot\dfrac{-9}{4}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{-3?}{16}=-\dfrac{3}{16}\)
\(\Leftrightarrow?=1\)
Mk gọi ? = x nha
a) \(\dfrac{3}{16}:\dfrac{x}{8}=\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{3}{16}:\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{1}{4}\)
⇒\(x=\dfrac{1.8}{4}=2\)
b) \(\dfrac{1}{25}:\dfrac{-3}{x}=\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{1}{25}:\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{-3}{5}\)
⇒x=5
c) \(\dfrac{x}{12}:\dfrac{-4}{9}=\dfrac{-3}{16}\)
\(\dfrac{x}{12}=\dfrac{-3}{16}.\dfrac{-4}{9}\)
\(\dfrac{x}{12}=\dfrac{1}{12}\)
⇒x=1
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)
b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)
c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)
c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{99.101}{100^2}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
Ta có: \(S=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=\dfrac{3}{2^2}\cdot\dfrac{2^3}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99}{10^2}\)
\(=\dfrac{11}{20}\)
bạn có thể giải thích rõ tại sao S=\(\dfrac{11}{20}\) đc ko
\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)
Từ đó ta có:
\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{50^2-1}{50^2}>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+...+1-\dfrac{1}{49.50}\)
\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)
\(\Rightarrow A>49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)
Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)
\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)