Đáp án: $x=\dfrac{\dfrac3{97}+\dfrac5{95}+\dfrac1{99}-\dfrac2{98}-\dfrac4{96}-\dfrac{968}{975}-\dfrac{6}{94}}{\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}-\dfrac{1}{975}-\dfrac{1}{94}}$
Giải thích các bước giải:
Ta có:
$\dfrac{x-1}{99}+\dfrac{x-3}{97}+\dfrac{x-5}{95}=\dfrac{x-2}{98}+\dfrac{x-4}{96}+\dfrac{x-968}{975}+\dfrac{x-6}{94}$
$\to \dfrac{x}{99}-\dfrac1{99}+\dfrac{x}{97}-\dfrac3{97}+\dfrac{x}{95}-\dfrac5{95}=\dfrac{x}{98}-\dfrac2{98}+\dfrac{x}{96}-\dfrac4{96}+\dfrac{x}{975}-\dfrac{968}{975}+\dfrac{x}{94}-\dfrac{6}{94}$
$\to \dfrac{x}{99}+\dfrac{x}{97}+\dfrac{x}{95}-\dfrac{x}{98}-\dfrac{x}{96}-\dfrac{x}{975}-\dfrac{x}{94}=\dfrac3{97}+\dfrac5{95}+\dfrac1{99}-\dfrac2{98}-\dfrac4{96}-\dfrac{968}{975}-\dfrac{6}{94}$
$\to x(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}-\dfrac{1}{975}-\dfrac{1}{94})=\dfrac3{97}+\dfrac5{95}+\dfrac1{99}-\dfrac2{98}-\dfrac4{96}-\dfrac{968}{975}-\dfrac{6}{94}$
$\to x=\dfrac{\dfrac3{97}+\dfrac5{95}+\dfrac1{99}-\dfrac2{98}-\dfrac4{96}-\dfrac{968}{975}-\dfrac{6}{94}}{\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}-\dfrac{1}{975}-\dfrac{1}{94}}$
