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\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)\)
\(=\left(1+1+1\right)+\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}\right)+\left(1+1+1+...+1\right)\)
\(=3+\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}\right)+\left(1+1+1+...+1\right)\)
Có: \(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}>0\)
\(1+1+1+...+1>0\)
=> \(3+\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}+...+\frac{2}{n^2+3n}\right)+\left(1+1+1+...+1\right)>3\)
Hay \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{n^2+3n}\right)>3\)
\(H=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow H=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\cdot\cdot\cdot\cdot\frac{99}{100}\)
\(\Leftrightarrow H=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)
\(\Leftrightarrow H=\frac{1}{100}\)
Ta có :
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)
\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1.2.2.3.3.4.....99.100}\)
\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1^2.2^2.3^2.4^2.....99^2}.\frac{1}{100}\)
\(=\)\(\frac{1}{100}\)
\(S=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)......\left(1-\frac{1}{100^2}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}\)
\(=\frac{\left(1.3\right).\left(2.4\right).........\left(99.101\right)}{\left(2.2\right).\left(3.3\right).......\left(100.100\right)}\)
\(=\frac{\left(1.2....99\right)\left(3.4....101\right)}{\left(2.3.....100\right)\left(2.3....100\right)}\)
\(=\frac{1.101}{100.2}=\frac{101}{200}\)
x-[17/2-6/35]=-1/3
x-583/70=-1/3
x=-1/3+583/70
x=1679/210
vậy x=1769/210
[2/3-(x-7/4)]=9/2+5/4
[2/3-(x-7/4)]=23/4
(x-7/4)=23/4+2/3
(x-7/4)=77/12
x=77/12+7/4
x=49/6
vậy x=49/6