Question

[Quang Vũ]

giải phương trình: 1/x + 5/2x+3 = 3/2x+1 + 4/x+3

Answer 1

[Akai Haruma]

Lời giải:
ĐKXĐ: $x\neq 0; \frac{-3}{2}; \frac{-1}{2}; -3$

PT $\Leftrightarrow (\frac{1}{x}-\frac{3}{2x+1})+(\frac{5}{2x+3}-\frac{4}{x+3})=0$

$\Leftrightarrow \frac{1-x}{x(2x+1)}+\frac{3-3x}{(2x+3)(x+3)}=0$

$\Leftrightarrow \frac{1-x}{x(2x+1)}+\frac{3(1-x)}{(2x+3)(x+3)}=0$

$\Leftrightarrow (1-x)\left[\frac{1}{x(2x+1)}+\frac{3}{(2x+3)(x+3)}\right]=0$

TH1: $1-x=0\Leftrightarrow x=1$ (tm) 

TH2: $\frac{1}{x(2x+1)}+\frac{3}{(2x+3)(x+3)}=0$

$\Rightarrow (2x+3)(x+3)+3x(2x+1)=0$

$\Leftrightarrow 8x^2+12x+9=0$

$\Leftrightarrow (2x+3)^2+4x^2=0$

$\Rightarrow (2x+3)^2=x^2=0$ (vô lý) 

Do đó $x=1$ là nghiệm duy nhất.