Đáp án:

$x=\pm\dfrac12$.

Giải thích các bước giải:

$x^2-\dfrac{14}{260}-\dfrac{14}{416}\ -\,.\!.\!.-\ \dfrac{14}{19400}=\dfrac{1}{25}\\\Rightarrow x^2-\left(\dfrac{14}{260}+\dfrac{14}{416}\ +\,.\!.\!.+\ \dfrac{14}{19400}\right)=\dfrac{1}{25}\\\Rightarrow x^2-\left(\dfrac{7}{130}+\dfrac{7}{208}\ +\,.\!.\!.+\ \dfrac{7}{9700}\right)=\dfrac{1}{25}\\\Rightarrow x^2-\left(\dfrac{7}{10.13}+\dfrac{7}{13.16}\ +\,.\!.\!.+\ \dfrac{7}{97.100}\right)=\dfrac{1}{25}\\\Rightarrow x^2-7.\left(\dfrac{1}{10.13}+\dfrac{1}{13.16}\ +\,.\!.\!.+\ \dfrac{1}{97.100}\right)=\dfrac{1}{25}\\\Rightarrow x^2-\dfrac73\cdot\left(\dfrac{3}{10.13}+\dfrac{3}{13.16}\ +\,.\!.\!.+\ \dfrac{3}{97.100}\right)=\dfrac{1}{25}\\\Rightarrow x^2-\dfrac73\cdot\left(\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\ +\,.\!.\!.+\ \dfrac{1}{97}-\dfrac{1}{100}\right)=\dfrac{1}{25}\\\Rightarrow x^2-\dfrac73\cdot\left(\dfrac{1}{10}-\dfrac{1}{100}\right)=\dfrac{1}{25}\\\Rightarrow x^2-\dfrac73\cdot\dfrac{9}{100}=\dfrac1{25}\\\Rightarrow x^2-\dfrac{21}{100}=\dfrac{1}{25}\\\Rightarrow x^2=\dfrac14\\\Rightarrow x=\pm\dfrac12$

Vậy $x=\pm\dfrac12$.

Đáp án:

Giải thích các bước giải:

`x^2 - 14/260 - 14/416 -..........- 14/19400 = 1/25`

`x^2 - 7/130 - 7/208 -.............- 7/9700 = 1/25`

`x^2 - ( 7/10.13 + 7/13.16 +........+ 7/97.100 ) = 1/25`

`x^2 - [ 7/3 . ( 1/10 - 1/13 + 1/13 - 1/16 +.........+ 1/97 - 1/100 )] = 1/25`

`x^2 - [ 7/3 . ( 1/10 - 1/100 ) ] = 1/25`

`x^2 - 21/100 = 1/25`

`x^2 = 4/100 + 21/100`

`x^2 = 25/100`

-Trường hợp chưa rút gọn:

`x^2 = (±5/10)^2`

`=> x in { 5/10; -5/10 }` 

-Trường hợp đã rút gọn:

`x^2 = 25/100`

`x^2 = 1/4`

`x^2 = (±1/2)^2`

`=> x in { 1/2; -1/2 }`