Giải thích các bước giải:
Ta có :
$(2x+1)^3+(x-2)^3=(3x-1)^3$
$\to (2x+1+x-2)((2x+1)^2-(2x+1)(x-2)+(x-2)^2)=(3x-1).(3x-1)^2$
$\to (3x-1)(3x^2+3x+7)=(3x-1).(9x^2-6x+1)$
$\to (3x-1).(9x^2-6x+1)-(3x-1)(3x^2+3x+7)=0$
$\to (3x-1).(6x^2-9x-6)=0$
$\to (3x-1).(2x^2-3x-2)=0$
$\to (3x-1) (x-2)(2x+1)=0$
$\to x\in\{\dfrac13,2,-\dfrac12\}$
Ta có :
$\dfrac{1}{x^2-1}+\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}=\dfrac{3}{7}$
$\to\dfrac{1}{(x-1)(x+1)}+\dfrac{1}{(x+1)(x+3)}+\dfrac{1}{(x+3)(x+5)}=\dfrac{3}{7}$
$\to\dfrac{2}{(x-1)(x+1)}+\dfrac{2}{(x+1)(x+3)}+\dfrac{2}{(x+3)(x+5)}=\dfrac{6}{7}$
$\to\dfrac{x+1-(x-1)}{(x-1)(x+1)}+\dfrac{x+3-(x+1)}{(x+1)(x+3)}+\dfrac{x+5-(x+3)}{(x+3)(x+5)}=\dfrac{6}{7}$
$\to\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}=\dfrac67$
$\to\dfrac{1}{x-1}-\dfrac{1}{x+5}=\dfrac67$
$\to (x+5)-(x-1)=\dfrac67(x+5)(x-1)$
$\to 6=\dfrac{6}{7}(x^2+4x-5)$
$\to 1=\dfrac{1}{7}(x^2+4x-5)$
$\to 7=x^2+4x-5$
$\to x^2+4x-12=0$
$\to (x-2)(x+6)=0$
$\to x\in\{-6,2\}$
