\(Tính: (1-\frac{1}{1+2}).(1-\frac{1}{1+2+3}).(1-\frac{1}{1+2+3+4})...(1-\frac{1}{1+2+3+...+20})\)
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Ta có: \(1-\frac{1}{1+2+...+n}=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}=1-2\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{2}{n}+\frac{2}{n+1}\) (*)
Áp dụng (*) vào bài toán ta được:
\(A=1-\frac{2}{2}+\frac{2}{3}+1-\frac{2}{3}+\frac{2}{4}+...+1-\frac{2}{20}+\frac{2}{21}\)
\(=1+1+...+1\left(19cs1\right)-\frac{2}{2}+\frac{2}{3}-\frac{2}{3}+\frac{2}{4}-\frac{2}{4}+...+\frac{2}{20}-\frac{2}{20}+\frac{2}{21}\)
\(=19-1+0+\frac{2}{21}=\frac{380}{21}\)

\(\Rightarrow B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+....+\frac{1}{20}.\frac{\left(1+20\right).20}{2}\)
\(\Rightarrow B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+...+\frac{1}{20}.\frac{21.20}{2}\)
\(\Rightarrow B=1+\frac{1}{2}.3+\frac{4}{2}+...+\frac{21}{2}\)
\(\Rightarrow B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(\Rightarrow B=\frac{2+3+4+...+21}{2}=...\)
Good Clever
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+...+\frac{1}{20}\cdot\frac{20\cdot21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{1+2+3+....+21}{2}-\frac{1}{2}\)
\(=\frac{21\cdot22}{2}\cdot\frac{1}{2}-\frac{1}{2}\)
\(=\frac{1}{2}\left(\frac{21\cdot22}{2}-1\right)\)
\(=230\cdot\frac{1}{2}\)
Bí

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)