Question

\(\dfrac{x}{x+1}-\dfrac{x-1}{3x+1}=\dfrac{1}{2}\)

Answer 1

ĐKXĐ: x\(\ne-1\), x\(\ne\dfrac{-1}{3}\)

\(\dfrac{x}{x+1}-\dfrac{x-1}{3x+1}=\dfrac{1}{2}\)

<=> \(\dfrac{\left(3x+1\right).2x}{\left(x+1\right)\left(3x+1\right).2}-\dfrac{\left(x-1\right)\left(x+1\right).2}{\left(x+1\right)\left(3x+1\right).2}=\dfrac{\left(x+1\right)\left(3x+1\right)}{\left(x+1\right)\left(3x+1\right).2}\)

=> 6x²+2x-2x²+2=3x²+x+3x+1

<=> 6x²+2x-2x²+2-3x²-x-3x-1=0

<=> x²-2x+1=0

<=> (x-1)²=0

<=> x-1=0

<=> x=1 (thỏa mãn ĐKXĐ)

Vậy S=\(\left\{1\right\}\)