Đáp án: $x=-200$

Giải thích các bước giải:

 Ta có:
$\dfrac{x+3}{197}+\dfrac{x+2}{198}+\dfrac{x+1}{199}=\dfrac{x+99}{101}+\dfrac{x+98}{102}+\dfrac{x+97}{103}$
$\to (1+\dfrac{x+3}{197})+(1+\dfrac{x+2}{198})+(1+\dfrac{x+1}{199})=(1+\dfrac{x+99}{101})+(1+\dfrac{x+98}{102})+(1+\dfrac{x+97}{103})$
$\to \dfrac{x+200}{197}+\dfrac{x+200}{198}+\dfrac{x+200}{199}=\dfrac{x+200}{101}+\dfrac{x+200}{102}+\dfrac{x+200}{103}$
$\to (x+200)\cdot (\dfrac1{197}+\dfrac1{198}+\dfrac1{199})=(x+200)\cdot (\dfrac1{101}+\dfrac1{102}+\dfrac1{103})$
$\to (x+200)\cdot (\dfrac1{197}+\dfrac1{198}+\dfrac1{199})-(x+200)\cdot (\dfrac1{101}+\dfrac1{102}+\dfrac1{103})=0$
$\to (x+200)\cdot ((\dfrac1{197}+\dfrac1{198}+\dfrac1{199})-(\dfrac1{101}+\dfrac1{102}+\dfrac1{103}))=0$
$\to x+200=0$
$\to x=-200$