Question

Giải phương trình ẩn x sau đây:

\(\dfrac{5x+3}{x-1}\) + \(\dfrac{3x}{x+1}\) = \(\dfrac{9x-4}{\left(x-1\right).\left(x+1\right)}\)

Answer 1

\(ĐK:x\ne\pm1\)

\(\dfrac{5x+3}{x-1}+\dfrac{3x}{x+1}=\dfrac{9x-4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{\left(5x+3\right)\left(x+1\right)+3x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{9x-4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\left(5x+3\right)\left(x+1\right)+3x\left(x-1\right)=9x-4\)

\(\Leftrightarrow5x^2+5x+3x+3+3x^2-3x-9x+4=0\)

\(\Leftrightarrow8x^2-4x+7=0\)

Vậy pt vô nghiệm

Answer 2

\(\Leftrightarrow\left(5x+3\right)\left(x+1\right)+3x\left(x-1\right)=9x-4\)

\(\Leftrightarrow5x^2+5x+3x+3+3x^2-3x-9x+4=0\)

\(\Leftrightarrow8x^2-4x+7=0\)

\(\text{Δ}=\left(-4\right)^2-4\cdot8\cdot7=-208< 0\)

Do đó: Phương trình vô nghiệm