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Từ GT ; ta có : \(\left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=224\)
\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=224\)
\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=224\)
\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=224\)
\(\Rightarrow\left(x-1\right).\dfrac{2}{9}=224\)
\(\Rightarrow\left(x-1\right)=1008\)
\(\Rightarrow x=1009\)
Vậy ...
Gọi biểu thức đó là A
\(A=\dfrac{1}{x}.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+..+\dfrac{1}{49.50}\right)=1\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)=1\)
Ta có công thức : \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức trên ta có :
\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)
......................
\(\dfrac{1}{49.50}=\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)
\(\Leftrightarrow A=\dfrac{1}{x}.\left(\dfrac{12}{25}\right)=1\)
\(\Rightarrow\) \(\dfrac{1}{x}=A:\dfrac{12}{25}=1:\dfrac{12}{25}=\dfrac{25}{12}\)
Vậy x = 12.
Mink nghĩ vậy, ai thấy đúng thì ủng hộ mink nha !!!
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
\(\frac{20}{9}-x=\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{72}\)
\(\Rightarrow\frac{20}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{8.9}\)
\(\Rightarrow\frac{20}{9}-x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+........+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow\frac{20}{9}-x=\frac{1}{3}-\frac{1}{9}\)
\(\Rightarrow\frac{20}{9}-x=\frac{2}{9}\)
\(\Rightarrow x=\frac{20}{9}-\frac{2}{9}\)
\(\Rightarrow x=\frac{18}{9}=2\)
20/9 -x= 1/3.4 +1/4.5+1/5.6+ 1/6.7+1/7.8+1/8.9
=1/3+1/4+1/4+1/5+1/5+1/6+1/6+1/7+1/7+1/8+1/8+1/9+1/9
20/9 -x= 1/3
x=20/9-1/3
x= 17/9
Nhớ tích (đúng) cho mình nha
\(\frac{20}{9}x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(\frac{20}{9}x=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(\frac{20}{9}x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{9}\)
\(\frac{20}{9}x=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
\(x=\frac{2}{9}:\frac{20}{9}=\frac{1}{10}\)
20/9 . 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 = 2/9
x = 2/9 : 20/9 = 2/9 . 9/20 = 1/10
Vậy x = 1/10
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
\(\dfrac{1}{6}x+\dfrac{1}{12}x+\dfrac{1}{20}x+...+\dfrac{1}{2450}x=1\)
\(x\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2450}\right)\)=1
\(x\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\right)\)=1
\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)
\(x\times\)\(\dfrac{12}{25}=1\)
\(\Rightarrow x=1\div\dfrac{12}{25}\)
\(x=1\times\dfrac{25}{12}=\dfrac{25}{12}\)
vậy \(x=\dfrac{25}{12}\)
vậy \(x=2\)\(x=2\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)\(\left(2\dfrac{2}{9}-x\right)\)=\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)
\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}\)