`x^2(2x+3)=2(3x-2)`

`<=>2x^3+3x^2=6x-4`

`<=>x^3+3/2x^2-3x+2=0(1)`

`a=1,b=3/2,c=-3,d=2`

Quy ước:

`\Delta=b^2-3ac`

`k=(9abc-2b^3-27a^2d)/(2\sqrt{|\Delta|^3})`

`=>\Delta(1)=(3/2)^2-3*1*(-3)`

`\Delta=9/4+9=45/4>0`

`k=(9*1*3/2*(-3)-2(3/2)^3-27*1*2)/(2\sqrt{45^3/4^3})`

`=>|k|=|(-405/4)/(2\sqrt{91125/64})| ~~1,341>1`

`\Delta>0,|k|>1`

`=>` phương trình có nghiệm duy nhất:

`x=(\sqrt{\Delta}|k|)/(3ak)(\root{3}{|k|+\sqrt{k^2-1}}+\root{3}{|k|-\sqrt{k^2-1}})-b/(3a)`

`x~~(3,35*1,31)/(-3,93)*2,06-1/2`

`x~~-2,81`

Đáp án:

Giải thích các bước giải:

`x^2(2x+3)=2(3x-2)`

$⇔2x^3+3x^2=6x-4$

$⇔2x^3+3x^2-6x+4=0$

`⇔x^3+3/2x^2-3x+2=0`

Đặt `x=y-1/2` thay vào phương trình mới:

$⇔\left(y-\dfrac{1}{2}\right)^3+\dfrac{3}{2}\cdot \left(y-\dfrac{1}{2}\right)^2-3\cdot \left(y-\dfrac{1}{2}\right)+2=0$

$⇔y^3-\dfrac{3y^2}{2}+\dfrac{3y}{4}-\dfrac{1}{8}+\dfrac{3}{2}\left(y^2-y+\dfrac{1}{4}\right)-3\left(y-\dfrac{1}{2}\right)+2$

$⇔y^3-\dfrac{3y^2}{2}+\dfrac{3y}{4}-\dfrac{1}{8}+\dfrac{3y^2-3y}{2}+\dfrac{3}{8}-3y+\dfrac{3}{2}+2=0$

$⇔\dfrac{4y^3-15y+15}{4}=0$

`⇔y^3-15/4y+15/4=0`

$\Delta'=\dfrac{\bigg(\dfrac{15}{4}\bigg)^2}{4}+\dfrac{\bigg(-\dfrac{15}{4}\bigg)^3}{27}=\dfrac{25}{16}>0$ 

`=>` $y=\sqrt[3]{\dfrac{\bigg(-\dfrac{15}{4}\bigg)}{2}+\sqrt{\dfrac{25}{16}}}+\sqrt[3]{\dfrac{\bigg(-\dfrac{15}{4}\bigg)}{2}-\sqrt{\dfrac{25}{16}}}=2,316996842$ 

Mà `x=y-1/2=>x=2,316996842-1/2=2,816996842`

Vậy nghiệm của phương trình là: `x=2,816996842`