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\(\frac{3}{4}\)*\(\frac{8}{9}\)*\(\frac{15}{16}\)********\(\frac{9999}{10000}\)
= \(\frac{1\cdot3}{2^2}\)*\(\frac{2\cdot4}{3^2}\)********\(\frac{99\cdot101}{100^2}\)
= \(\frac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\)* \(\frac{3\cdot4\cdot5\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot100}\)
= \(\frac{1}{100}\)*\(\frac{101}{2}\)=\(\frac{101}{200}\)
Ta có: A = \(\frac{3}{8}\). \(\frac{8}{9}\).\(\frac{15}{16}\). ... .\(\frac{9999}{10000}\)
\(\Rightarrow\) A = \(\frac{1.3}{2^2}\).\(\frac{2.4}{3^2}\). \(\frac{3.5}{4^2}\). ... . \(\frac{99.101}{100^2}\)
\(\Rightarrow\) A = \(\frac{1.111}{2.100}\)= \(\frac{111}{200}\)
Vậy: A = \(\frac{111}{200}\).
\(8\frac{4}{17}-\left(2\frac{5}{9}+3\frac{4}{17}\right)=\frac{140}{17}-\left(\frac{23}{9}+\frac{55}{17}\right)=\frac{140}{17}-\frac{886}{153}=\frac{22}{9}=2,444444444444\)
\(a)\) Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(............\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\)\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\)\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\)\(A< 1+1-\frac{1}{100}\)
\(\Rightarrow\)\(A< 2-\frac{1}{100}< 2\)
\(\Rightarrow\)\(A< 2\) ( đpcm )
Vậy \(A< 2\)
Chúc bạn học tốt ~
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\frac{1}{2}.\left(\frac{20}{15}+\frac{6}{15}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{48}{9}\right)\)
\(=\frac{1}{2}.\frac{26}{15}-\frac{3}{4}.\frac{56}{9}\)
\(=\frac{13}{15}-\frac{14}{3}\)
\(=-\frac{19}{5}\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\left(\frac{1}{2}.\frac{4}{3}+\frac{1}{2}.\frac{2}{5}\right)-\left(\frac{3}{4}.\frac{8}{9}+\frac{3}{4}.\frac{16}{3}\right)\)
\(=\left(\frac{2}{3}+\frac{1}{5}\right)-\left(\frac{2}{3}+4\right)\)
\(=\frac{2}{3}+\frac{1}{5}-\frac{2}{3}-4\)
\(=\frac{1}{5}-4\)
\(=\frac{1}{5}-\frac{20}{5}=\frac{-19}{5}\)
Ta thấy : \(4=2^2;9=3^2;....;10000=100^2\) nên A có \(\left(100-2\right):1+1=99\) số hạng
Ta có :
\(\frac{3}{4}< \frac{4}{4}=1\)
\(\frac{8}{9}< \frac{9}{9}=1\)
\(\frac{15}{16}< \frac{16}{16}=1\)
\(......\)
\(\frac{9999}{10000}< \frac{10000}{10000}=1\)
\(\Rightarrow A=\frac{3}{4}+\frac{8}{9}+....+\frac{9999}{10000}< 1+1+...+1\)(Vì A có 99 số hạng nên cũng có 99 số 1 tương ứng)
\(\Rightarrow A< 99\)
\(A=\frac{3}{4}+\frac{8}{9}+...+\frac{9999}{10000}\)
\(A=1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{10000}\)
\(A=99-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)\)
Vì biểu thức trong dấu ngoặc đơn luôn lớn hơn 0 nên A<99
Vậy A<99
\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+-\frac{37}{100}+\frac{1}{8}+-\frac{128}{100}+-\frac{250}{100}+\frac{37}{12}\)
\(=\left(\frac{19}{4}+\frac{1}{8}+\frac{37}{12}\right)-\left(\frac{37}{100}+\frac{128}{100}+\frac{250}{100}\right)\)
\(=\left(\frac{114}{24}+\frac{3}{24}+\frac{74}{24}\right)-\frac{415}{100}\)
\(=\frac{191}{24}-\frac{415}{100}\)
\(=\frac{457}{120}\)
Tham khảo nha !!!
\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+\frac{-37}{100}+\frac{1}{8}+\frac{-32}{25}+\frac{-5}{2}+\frac{37}{12}\)
\(=\left(\frac{19}{4}+\frac{1}{8}+\frac{-5}{2}\right)+\left(\frac{-37}{100}+\frac{-32}{25}\right)+\frac{37}{12}\)
\(=\frac{19}{8}+\frac{-33}{20}+\frac{37}{12}\)
\(=\frac{29}{40}+\frac{37}{12}\)
\(=\frac{457}{120}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{3.8.15...9999}{4.9.16...10000}=\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}=\frac{\left(1.2.3...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(\frac{1.101}{100.2}=\frac{101}{200}\)
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mình nhầm chút nha.nếu là so sánh thì mình làm được