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4, $(2x-5)^2- 9(x+1)^2 = 0$
$\leftrightarrow (2x-5-3x-3)(2x-5+3x+3)=0$
$\leftrightarrow (-x-8)(5x-2)=0$
$\leftrightarrow x=-8$ hoặc $x=\dfrac{2}{5}$
3, $\dfrac{6x+8}{16x^2- 1}-\dfrac{2}{4x+1}=\dfrac{3}{4x-1}$ (ĐKXĐ: $x \neq \pm \; \dfrac{1}{4})$
$\leftrightarrow \dfrac{6x+8}{(4x+1)(4x-1)}-\dfrac{2(4x-1)}{(4x+1)(4x-1)}=\dfrac{3(4x+1)}{(4x+1)(4x-1)}$
$\rightarrow 6x+8-8x+2=12x+3$
$\leftrightarrow -14x=-7$
$\leftrightarrow x=\dfrac{1}{2}$
2, $\dfrac{2}{x-1}+\dfrac{3}{x-2}=\dfrac{3}{x-3}$ (ĐKXĐ: $x \neq 1 ; x \neq 2 ; x \neq 3)$
$\rightarrow 2(x-2)(x-3)+3(x-1)(x-3)=3(x-1)(x-2)$
$\leftrightarrow 5x^2-22x+21=3x^2-9x+6$
$\leftrightarrow (x-5)(2x-3)=0$
$\leftrightarrow x=5$ hoặc $x=\dfrac{3}{2}$
1, Mình chưa dịch được đề :| gõ latex bạn nhé