Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Link này bạn:Câu hỏi của Hoàng Hà Nhi - Toán lớp 6 | Học trực tuyến
\(\dfrac{x-1}{12}+\dfrac{x-1}{20}+\dfrac{x-1}{30}+...+\dfrac{x-1}{72}=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{8.9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\dfrac{2}{9}=\dfrac{16}{9}\\ x-1=8\\ x=8+1\\ x=9\)
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
ta có
x-1/12+x-1/20+x-1/30+x-1/42+x-1/56+x-1/72=16/9
=>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(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9)=16/9
=>x-1*(1/3-1/9)=16/9
=>(x-1)*2/9=16/9
=>x-1=9
=>x=8
kb và like cho mình nhé 
a,\(2\frac{2}{9}x=\frac{1}{12}+\frac{1}{20}+............+\frac{1}{72}\)
=>\(\frac{20}{9}x=\frac{1}{3.4}+\frac{1}{4.5}+.............+\frac{1}{8.9}\)
=>\(\frac{20}{9}x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.............+\frac{1}{8}-\frac{1}{9}\)
=>\(\frac{20}{9}x=\frac{1}{3}-\frac{1}{9}\)
=>\(\frac{20}{9}x=\frac{2}{9}\)
=>x=\(\frac{1}{10}\)
b,\(\left(\frac{1}{2.3}+\frac{1}{3.4}+.............+\frac{1}{45.50}\right)x=1\)
=>\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{45}-\frac{1}{50}\right)x=1\)
=>\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
=>\(\frac{12}{25}x=1\)
=>\(x=\frac{25}{12}\)
\(\frac{20}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
\(\Rightarrow\frac{20}{9}-x=\frac{2}{9}\Rightarrow x=\frac{20}{9}-\frac{2}{9}=\frac{18}{9}=2\)
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
\(2\dfrac{2}{9}-x=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3}-\dfrac{1}{9}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{2}{9}\)
\(\Rightarrow x=2\dfrac{2}{9}-\dfrac{2}{9}\)
\(\Rightarrow x=2\)
Vậy x=2
2\(\dfrac{2}{9}\) - x = \(\dfrac{1}{3\cdot4}\)+\(\dfrac{1}{4\cdot5}\)+\(\dfrac{1}{5\cdot6}\)+\(\dfrac{1}{6\cdot7}\)+\(\dfrac{1}{7\cdot8}\)+\(\dfrac{1}{8\cdot9}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{1}{3}\)-\(\dfrac{1}{9}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{9}{27}\)- \(\dfrac{3}{27}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{2}{9}\)
\(\dfrac{20}{9}\) -x = \(\dfrac{2}{9}\)
x = \(\dfrac{20}{9}-\dfrac{2}{9}\)
x = 2
Vậy x = 2
6x-(1/12+1/20+1/30+1/42+1/56+1/72)=224
6x-(1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9)=224
6x-(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9)=224
6x-(1/3-1/9)=224
6x-2/9=224
6x=2018/9
x=1009/27