Đáp án: $x=\dfrac{3\pm\sqrt{37}}{2}$
Giải thích các bước giải:
Ta có:
$2(x^2+2x+3)=5\sqrt{x^3+3x^2+3x+2}$
$\to 2((x+1)^2+2)=5\sqrt{(x+1)^3+1}$
Đặt $(x+1)=y$
$\to 2(y^2+2)=5\sqrt{y^3+1}$
$\to 2y^2+4=5\sqrt{y^3+1}$
$\to (2y^2+4)^2=25(y^3+1)$
$\to 4y^4+16y^2+16=25y^3+25$
$\to 4y^4-25y^3+16y^2-9=0$
$\to (4y^2-5y+3)(y^2-5y-3)=0$
$\to y^2-5y-3=0$ vì $4y^2-5y+3=4(y-\dfrac58)^2+\dfrac{23}{16}$
$\to y=\dfrac{5\pm\sqrt{37}}{2}$
$\to x+1=\dfrac{5\pm\sqrt{37}}{2}$
$\to x=\dfrac{5\pm\sqrt{37}}{2}-1$
$\to x=\dfrac{3\pm\sqrt{37}}{2}$
