[toán 9] violimpic vòng 12

H

huynhbachkhoa23

$(x+1)(x+3)-(x+3)\sqrt{x^2+1}=0$

$\leftrightarrow (x+3)(x+1-\sqrt{x^2+1})=0$

$\rightarrow x=-3; x=0$

Vậy $S=-3$
 
T

thinhrost1

gif.latex

Cộng các nghiệm lại
 
B

baochauhn1999

$x^2+4x+3=(x+3)\sqrt{x^2+1}$
ĐKXĐ: Tự tìm
$<=>(x+3)(x+1)-(x+3)\sqrt{x^2+1}=0$
$<=>(x+3)(x+1-\sqrt{x^2+1})=0$
$=>x_1=-3$
$x+1=\sqrt{x^2+1}$
$<=>x^2+2x+1=x^2+1$
$<=>2x=0$
$<=>x=0$
$=>S=-3$
 
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