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Theo mk được biết thì Shinichi và Kid là hai anh em nên mk thích cả hai
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xem ai thông minh, tinh mắt nhất có thể luận ra toàn bộ đề và giúp mk giải nào!!
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Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
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\(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{97.100}\)
\(=3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=3\left(1-\dfrac{1}{100}\right)\)
\(=3.\dfrac{99}{100}=\dfrac{297}{100}\)
Vậy...
\(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{97.100}\)
\(=3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=3\left(1-\dfrac{1}{100}\right)=3.\dfrac{99}{100}=\dfrac{297}{100}\)
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\(x=\dfrac{7}{25}+\dfrac{-1}{5}.\Leftrightarrow x=\dfrac{7-5}{25}=\dfrac{2}{25}.\)
\(x=\dfrac{3}{11}+\dfrac{4}{-9}.\Leftrightarrow x=\dfrac{33-44}{99}=\dfrac{-1}{9}.\)
\(\dfrac{5}{9}+\dfrac{x}{-1}=\dfrac{-1}{3}.\Leftrightarrow\dfrac{5}{9}-x=\dfrac{-1}{3}.\Leftrightarrow x=\dfrac{8}{9}.\)
\(a,\Rightarrow x=\dfrac{7}{25}+\dfrac{-5}{25}\\ \Rightarrow x=\dfrac{2}{25}\\ b,\Rightarrow x=\dfrac{5}{11}+\dfrac{-4}{9}\\ \Rightarrow x=\dfrac{45}{99}+\dfrac{-44}{99}\\ \Rightarrow x=\dfrac{1}{99}\)