Tìm x: \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\cdot\left(x+1\right)}=\)
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Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\)
=> \(A=\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x\left(x+1\right)}\)
\(\frac{A}{2}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)
=> \(\frac{A}{2}=\frac{1}{6}-\frac{1}{x+1}=\frac{x+1-6}{6\left(x+1\right)}=\frac{x-5}{6\left(x+1\right)}\) => \(A=\frac{x-5}{3\left(x+1\right)}=\frac{2}{9}\)
<=> 3(x-5)=2(x+1) <=> 3x-15=2x+2 <=> x=17
Đáp số: x=17
a/(x+1)+(x+2)+...+(x+100)=5750
x+1+x+2+...+x+100=5750
100x+(1+2+...+100)=5750
100x+50.(100+1)=5750
100x+5050=5750
100x=700
x=7.
b/ 1/1.2+1/2.3+...+1/x(x+1)=2015/2016
1/1-1/2+1/3-1/4+...+1/x-1/x+1=2015/2016
1-1/x+1=2015/2016
1/x+1=1/2016
x+1=2016
x=2015
\(\left[1-\frac{1}{21}\right]\times\left[1-\frac{1}{28}\right]\times\left[1-\frac{1}{36}\right]\times...\times\left[1-\frac{1}{1326}\right]\)
\(=\frac{20}{21}\times\frac{27}{28}\times\frac{35}{36}\times...\times\frac{1325}{1326}\)
\(=\frac{40}{42}\times\frac{54}{56}\times\frac{70}{72}\times...\times\frac{2650}{2652}\)
\(=\frac{5\times8}{6\times7}\times\frac{6\times9}{7\times8}\times\frac{7\times10}{8\times9}\times...\times\frac{50\times53}{51\times52}\)
\(=\frac{5\times6\times7\times...\times50}{6\times7\times8\times...\times51}\times\frac{8\times9\times10\times...\times53}{7\times8\times9\times...\times52}\)
\(=\frac{5}{51}\times\frac{53}{7}\)
\(=\frac{265}{357}\)
* ĐK: \(x\ne0\)
Đề ra ...<=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{1}{9}\)
<=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{6}-\frac{1}{x+1}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{x+1}\left(1-\frac{1}{x}\right)=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{x-1}{x\left(x+1\right)}=\frac{1}{36}\)
<=> \(\frac{x-1}{x\left(x-1\right)}=\frac{x-1}{36.\left(x-1\right)}\)
=> x(x-1) = 36. (x-1) => x =36
\(\frac{2}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x+\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
x+1=18
x=18-1
x=17
1/21 + 1/28 + 1/36 + ... + 2/x(x + 1) = 2/9
1/2 × (1/21 + 1/28 + 1/36 + ... + 2/x(x + 1) = 1/2 × 2/9
1/42 + 1/56 + 1/72 + ... + 1/x(x + 1) = 1/9
1/6×7 + 1/7×8 + 1/8×9 + ... + 1/x(x + 1) = 1/9
1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x + 1 = 1/9
1/6 - 1/x + 1 = 1/9
1/x + 1 = 1/6 - 1/9
1/x + 1 = 3/18 - 2/18
1/x + 1 = 1/18
=> x + 1 = 18
=> x = 18 - 1
=> x = 17