A=1/21+1/28+1/36+...+2/x(x+1)=2/9
1/21 + 1/28 + 1/36 +...+ 2/x.(x+1) = 2/9
1/21 + 1/28 + 1/36 +...+ 2/x.(x+1) = 2/9
1/21 + 1/28 + 1/36 +...+ 2/x.(x+1) = 2/9
1/21 + 1/28 + 1/36 +...+ 2/x.(x+1) = 2/9
1/21 + 1/28 + 1/36 +...+ 2/x.(x+1) = 2/9
1/21+1/28+1/36+...+2/x(x+1)=2/9
1/21+1/28+1/36+...+2/x(x+1)=2/9
=>1/3.7+1/4.7+1/4.9+...+2/x.(x+1)=2/9
=>2/2.3.7+2/2.4.7+2/2.4.9+...+2/x.(x+1)=2/9
=>2/6.7+2/7.8+2/8.9+...+2/x.(x+1)=2/9
=>2.(1/6-1/7+1/7-1/8+1/8-1/9+...+1/x-1/x-1)=2/9
=>2.(1/6-1/x+1)=2/9
=>1/6-1/x+1=1/9
=>1/x+1=1/6-1/9
=>1/x+1=1/18
=>x+1=18
=>x=18-1
=>x=17
1/21 + 1/28 +1/36 +... + 2/x(x+1) = 2/9
Ta có:
1/x - 1/(x+1) = (x + 1 - x)/x(x+1) = 1/x(x+1)
⇒ 2/x(x+1) = 2/x - 2/(x+1)
(*) trở thành:
2/6 - 2/7 + 2/7 - 2/8 + ... + 2/x - 2/(x+1) = 2/9
⇔ 2/6 - 2/(x+1) = 2/9
⇔ 1/6 - 1/(x+1) = 1/9
⇔ 1/(x+1) = 1/18
⇔ x + 1 = 18
⇔ x = 17
Đáp số: x = 17
1/21+1/28+1/36+...+2/x(x+1)=2/9
Ta có \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}\)
=\(2\left(\frac{1}{42}+\frac{1}{56}+....+\frac{1}{x\left(x+1\right)}\right)\)
=\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{x\left(x+1\right)}\right)\)
=\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}\right)\)
=\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)\)
=>\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
=>\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}:2\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
=> x+1=18
x=18-1
x=17
2 / 6,7 + 2 / 7.8 + ... + 2 / x (x + 1) = 2/9 Ta có: 1 / x - 1 / (x + 1) = (x + 1 - x) / x (x 1) = 1 / x (x + 1) ⇒ 2 / x (x + 1) = 2 / x - 2 / (x + 1) (*) trở thành: 2/6 - 2/7 + 2/7 - 2/8 + ... + 2 / x - 2 / (x + 1) = 2/9 ⇔ 2/6 - 2 / (x + 1) = 2/9 ⇔ 1/6 - 1 / (x + 1) = 1/9 ⇔ 1 / (x + 1) = 1/18 ⇔ x + 1 = 18 ⇔ x = 17 Đáp số: x = 17
Tìm x: 1/21+1/28+1/36+...+2/x.(x+1)=2/9
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
2\(\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=> x+1 =18
=> x = 18 - 1
=> x = 17