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\(\frac{1}{1-\frac{2}{1-\frac{3}{1-\frac{1}{4}}}}=\frac{1}{1-\frac{2}{1-\frac{3}{\frac{3}{4}}}}=\frac{1}{1-\frac{2}{1-4}}=\frac{1}{1-\frac{2}{-3}}=\frac{1}{\frac{5}{3}}=\frac{3}{5}\Rightarrow A=1-\frac{3}{5}=\frac{2}{5}\)
Bài làm
\(A=1-\frac{1}{1-\frac{2}{1-\frac{3}{1-\frac{1}{4}}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-\frac{3}{\frac{4}{4}-\frac{1}{4}}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-\frac{3}{\frac{3}{4}}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-3:\frac{3}{4}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-4}}\)
\(A=1-\frac{1}{1-\frac{2}{-3}}\)
\(A=1-\frac{1}{1+\frac{2}{3}}\)
\(A=1-\frac{1}{\frac{3}{3}+\frac{2}{3}}\)
\(A=1-\frac{1}{\frac{5}{3}}\)
\(A=1-1:\frac{5}{3}\)
\(A=1-\frac{3}{5}\)
\(A=\frac{5}{5}-\frac{3}{5}\)
\(A=\frac{2}{5}\)
Vậy \(A=\frac{2}{5}\)
# Học tốt #
ta có:\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\)\(\Rightarrow\frac{1}{2}\times a\times\frac{1}{6}=\frac{2}{3}\times b\times\frac{1}{6}=\frac{3}{4}\times c\times\frac{1}{6}\)
\(\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)
\(\Rightarrow\frac{a}{12}=5\Rightarrow a=12\times5=60\)
\(\Rightarrow\frac{b}{9}=5\Rightarrow b=9\times5=45\)
\(\Rightarrow\frac{c}{8}=5\Rightarrow c=8\times5=40\)
chúc bạn học tốt!!
\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c=\frac{a}{2}=\frac{2b}{3}=\frac{3b}{4}\)
\(\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}=\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)
\(\Rightarrow a=5.12=60\); \(b=5.9=45\); \(c=5.8=40\)
Vậy \(a=60\), \(b=45\), \(c=40\)
\(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
=> \(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
=> \(2C=1-\frac{1}{3^{99}}\)
=> \(C=\frac{1-\frac{1}{3^{99}}}{2}\)
Vì\(1-\frac{1}{3^{99}}< 1\Rightarrow\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3B-B=1-\frac{1}{3^{99}}\)
\(B=\frac{1-\frac{1}{3^{99}}}{2}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(3A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
Chúc bạn học tốt ~
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
Đặt: \(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+........+2019}\)
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .............. ; \(1+2+3+......+2019=\frac{2019.2020}{2}\)
\(\Rightarrow B=\frac{2}{2}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+........+\frac{1}{\frac{2019.2020}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{2019.2020}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2019.2020}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2.\left(1-\frac{1}{2020}\right)=2.\frac{2019}{2020}=\frac{2019}{1010}\)
\(\Rightarrow A=\frac{2.2019}{\frac{2019}{1010}}=2.1010=2020\)
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