giúp mik với ạ :(
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Ta có:
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\)
\(\Leftrightarrow3A=\frac{3\cdot3}{1\cdot4}+\frac{3\cdot3}{4\cdot7}+\frac{3\cdot3}{7\cdot10}+...+\frac{3\cdot3}{97\cdot100}\)
\(\Leftrightarrow3A=\frac{9}{1\cdot4}+\frac{9}{4\cdot7}+\frac{9}{7\cdot10}+...+\frac{9}{97\cdot100}\)
\(\Leftrightarrow3A=\left(9-\frac{9}{4}\right)+\left(\frac{9}{4}-\frac{9}{7}\right)+\left(\frac{9}{7}-\frac{9}{10}\right)+...+\left(\frac{9}{97}-\frac{9}{100}\right)\)
\(\Leftrightarrow3A=9-\frac{9}{4}+\frac{9}{4}-\frac{9}{7}+\frac{9}{7}-\frac{9}{10}+...-\frac{9}{100}\)
\(\Leftrightarrow3A=9+\left(\frac{9}{4}-\frac{9}{4}\right)+\left(\frac{9}{7}-\frac{9}{7}\right)+\left(\frac{9}{10}-\frac{9}{10}\right)+...-\frac{9}{100}\)
\(\Leftrightarrow3A=9+0+0+0-\frac{9}{100}\)
\(\Leftrightarrow3A=9-\frac{9}{100}\)
\(\Leftrightarrow3A=\frac{9\cdot100-9}{100}\)
\(\Leftrightarrow3A=\frac{891}{100}\)
\(\Leftrightarrow A=\frac{891}{100}:3\)
\(\Leftrightarrow A=\frac{891}{100\cdot3}\)
\(\Leftrightarrow A=\frac{297\cdot3}{100\cdot3}\)
\(\Leftrightarrow A=\frac{297}{100}\)
Ta có:
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\)
\(\Leftrightarrow3A=\frac{3\cdot3}{1\cdot4}+\frac{3\cdot3}{4\cdot7}+\frac{3\cdot3}{7\cdot10}+...+\frac{3\cdot3}{97\cdot100}\)
\(\Leftrightarrow3A=\frac{9}{1\cdot4}+\frac{9}{4\cdot7}+\frac{9}{7\cdot10}+...+\frac{9}{97\cdot100}\)
\(\Leftrightarrow3A=\left(9-\frac{9}{4}\right)+\left(\frac{9}{4}-\frac{9}{7}\right)+\left(\frac{9}{7}-\frac{9}{10}\right)+...+\left(\frac{9}{97}-\frac{9}{100}\right)\)
\(\Leftrightarrow3A=9-\frac{9}{4}+\frac{9}{4}-\frac{9}{7}+\frac{9}{7}-\frac{9}{10}+...-\frac{9}{100}\)
\(\Leftrightarrow3A=9+\left(\frac{9}{4}-\frac{9}{4}\right)+\left(\frac{9}{7}-\frac{9}{7}\right)+\left(\frac{9}{10}-\frac{9}{10}\right)+...-\frac{9}{100}\)
\(\Leftrightarrow3A=9+0+0+0-\frac{9}{100}\)
\(\Leftrightarrow3A=9-\frac{9}{100}\)
\(\Leftrightarrow3A=\frac{9\cdot100-9}{100}\)
\(\Leftrightarrow3A=\frac{891}{100}\)
\(\Leftrightarrow A=\frac{891}{100}:3\)
\(\Leftrightarrow A=\frac{891}{100\cdot3}\)
\(\Leftrightarrow A=\frac{297\cdot3}{100\cdot3}\)
\(\Leftrightarrow A=\frac{297}{100}\)
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Ta có:
\(T=\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot100}\)
\(\Leftrightarrow2T=\frac{4\cdot2}{1\cdot3}+\frac{4\cdot2}{3\cdot5}+\frac{4\cdot2}{5\cdot7}+...+\frac{4\cdot2}{99\cdot100}\)
\(\Leftrightarrow2T=\frac{8}{1\cdot3}+\frac{8}{3\cdot5}+\frac{8}{5\cdot7}+...+\frac{8}{99\cdot100}\)
\(\Leftrightarrow2T=\left(\frac{8}{1}-\frac{8}{3}\right)+\left(\frac{8}{3}-\frac{8}{5}\right)+...+\left(\frac{8}{99}-\frac{8}{100}\right)\)
\(\Leftrightarrow2T=8-\frac{8}{3}+\frac{8}{3}-\frac{8}{5}+\frac{8}{5}-...+\frac{8}{99}-\frac{8}{100}\)
\(\Leftrightarrow2T=8+\left(\frac{8}{3}-\frac{8}{3}\right)+\left(\frac{8}{5}-\frac{8}{5}\right)+...-\frac{8}{100}\)
\(\Leftrightarrow2T=8+0+0+...-\frac{8}{100}\)
\(\Leftrightarrow2T=8-\frac{8}{100}\)
\(\Leftrightarrow2T=\frac{8\cdot100-8}{100}\)
\(\Leftrightarrow2T=\frac{792}{100}\)
\(\Leftrightarrow T=\frac{792}{100}:2\)
\(\Leftrightarrow T=\frac{792}{100\cdot2}\)
\(\Leftrightarrow T=\frac{198\cdot2\cdot2}{25\cdot2\cdot2}\)
\(\Leftrightarrow T=\frac{198}{25}\)
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Ta có \(F=2F-F=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2022}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2022}}\right)\)
\(=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2021}}-1-\frac{1}{2}-\frac{1}{2^2}-..-\frac{1}{2^{2022}}\)
\(=2-\frac{1}{2^{2022}}\)
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a) Tổ 1 và tổ 2 trồng được số phần so với số cây trồng được cả lớp là:
\(\frac{1}{3}+\frac{5}{12}=\frac{3}{4}\)
Tổ 3 trồng được số phần so với số cây cả lớp trồng được là:
\(1-\frac{3}{4}=\frac{1}{4}\)
Số cây cả lớp trồng được là:
\(30:\frac{1}{4}=120\left(cây\right)\)
Tổ 1 trồng được số cây là:
\(120×\frac{1}{3}=40\left(cây\right)\)
Tổ 2 trồng được số cây là:
\(120×\frac{5}{12}=50\left(cây\right)\)
b) Tỉ số phần trăm số cây tổ 1 và tổ 2 trồng được là:
\(\frac{40}{50}×100\%=80\%\)
Đáp số: a) Tổ 1: \(40cây\)
Tổ 2: \(50cây\)
Tổ 3: \(30cây\)
b) \(80\%\)
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