Đáp án:
$\dfrac{3x-1}{x-1}$ $-$ $\dfrac{2x+5}{x+3}$ $=$ $\dfrac{4}{(x-1)(x+3)}$ (Điều kiện $x$$\neq$$1$;$ x $$\neq$$-3$)
$⇔$ $\dfrac{(3x-1)(x+3)-(2x+5)(x-1)}{(x-1)(x+3)}$ $=$ $\dfrac{4}{(x-1)(x+3)}$
$⇔$ $\dfrac{3x^2+9x-x-3-(2x^2-2x+5x-5)}{(x-1)(x+3)}$ $=$ $\dfrac{4}{(x-1)(x+3)}$
$⇔ 3x²+9x-x-3-(2x²-2x+5x-5) = 4$
$⇔ 3x²+9x-x-3-2x²+2x-5x+5-4=0$
$⇔ x²+5x-2=0$
$⇔ x²+2.x.2,5+2,5²-8,25=0$
$⇔ (x+2,5)²-8,25=0$
$⇔ (x+2,5)²=8,25$
$⇔ x+2,5=√8,25$ hoặc $x+2,5=-√8,25$
$⇔ x= -2,5+√8,25$ (nhận) hoặc $x= -2,5-√8,25$ (nhận)
BẠN THAM KHẢO NHA!!!
