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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}...-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
"girl cute" là sai rồi bạn ơi, trong tiếng anh, tính từ (cute) phải đứng trước danh tư (girl).
![](https://rs.olm.vn/images/avt/0.png?1311)
2 2/9 - x = 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
20/9 - x = 1/3×4 + 1/4×5 + 1/5×6 + 1/6×7 + 1/7×8 + 1/8×9
20/9 - x = 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/8 - 1/9
20/9 - x = 1/3 - 1/9
20/9 - x = 3/9 - 1/9
20/9 - x = 2/9
x = 20/9 - 2/9
x = 18/9 = 2
Vậy x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{9}\right)=1+\frac{1}{2}-\frac{1}{3}=1\frac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:$\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}$
=> x+1(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
=> x+1.2/9=16/9
=> x+1 = (16/9):(2/9)
=> x+1 = 8
=> x = 9
thông cảm mình ko đánh được dấu ngoặc tròn
[x-1].[1/12+1/20+1/30+1/42+1/56+1/72] =16/9
[x-1].[1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9]=16/9
[x-1].[1/3-1/9]=16/9
[x-1].2/9=16/9
x-1=16/9:2/9
x-1=8
x=7
Vậy x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
đặt S=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(S=1-\frac{1}{9}=\frac{8}{9}\)
Vậy tổng=1/90+8/9=9/10
\(\frac{20}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
\(\Rightarrow\frac{20}{9}-x=\frac{2}{9}\Rightarrow x=\frac{20}{9}-\frac{2}{9}=\frac{18}{9}=2\)
Vậy x = 2.
k cho mk nha
\(2\frac{2}{9}-x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(2\frac{2}{9}-x=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{9}\)
\(2\frac{2}{9}-x=\frac{3}{9}-\frac{1}{9}\)
\(2\frac{2}{9}-x=\frac{2}{9}\)
\(x=2\frac{2}{9}-\frac{2}{9}\)
\(x=2\)
Vậy x = 2