\(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+...+\frac{7}{10^{2019}}\)
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đặt \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+\frac{7}{10^4}\)
\(A=7.\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
Lại đặt \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)
\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)
\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
\(9B=1-\frac{1}{10^4}\)
\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)
\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)
a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
Có:495.810/147.49.413=710.230/233.79=7/8 7/10-7/12+7/5 / 8/10-8/12+8/5=7(1/10-1/12+1/5) / 8(1/10-1/12+1/5)=7/8 =>A=7/8-7/8=0 xin lỗi nha mik ghi hơi khó hiểu chút vì mik mới dùng online math.HOK TỐT
\(A=\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{100}}\)
\(10A=7+\frac{7}{10}+...+\frac{7}{10^{99}}\)
\(\Rightarrow10A-A=9A=7-\frac{7}{10^{100}}\)
1/10 A =7/10^2+7/10^3+..............+7/10^2020
9/10*A=(7/10+7/10^2+......................+7/10^2019)-(7/10^2+7/10^3+........+7/10^2020)
=7/10-7/10^2020
A=10/9 .(7/10-7/10^2020)