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a,1/5+2/5+3/5+4/5+...+9/5
=(1+2+3+4+...+9)/5
=45/5
=9
b,17,8(3,7+5,7)-7,8(4,6+4,8)
=17,8.9,4-7,8.9,4
=9,4(17,8-7,8)
=9,4.10
=94
đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(2S=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(\Leftrightarrow2S-S=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(\Leftrightarrow S=\frac{1}{101}-1=-\frac{100}{101}\)
\(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{2018.2023}\)
Ta có : \(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{2018.2023}\)
\(=\frac{1}{5}.\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{2018.2023}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{2018}-\frac{1}{2023}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{2023}\right)\)
\(=\frac{1}{5}.\frac{2020}{6069}=\frac{404}{6069}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(< =>2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(< =>2A-A=1-\frac{1}{2^{99}}< =>A=1-\frac{1}{2^{99}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{99}}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}\)
khỏi ghi lại đề nha
A=1-1/2+1/2-1/3+1/3-1/4+......+1/49-1/50
A=1-1/50
A=49/50
\(a,\frac{15}{2}-\left(\frac{x}{2}-\frac{3}{4}\right)=\frac{5}{26}\)
\(\frac{x}{2}-\frac{3}{4}=\frac{15}{2}-\frac{5}{26}\)
\(\frac{x}{2}-\frac{3}{4}=39\)
\(\frac{x}{2}=39+\frac{3}{4}\)
\(\frac{x}{2}=\frac{159}{4}\)
\(\Rightarrow\frac{2.x}{4}=\frac{159}{4}\)
\(\Rightarrow2.x=159\)
\(\Rightarrow x=159:2=\frac{159}{2}\)
232
\(\frac{32}{99}\)