Ta có : \(\left(x+2\right)^3+\left(3-x\right)^3=-18\)
\(\Leftrightarrow\left(x+2+3-x\right)\left(x^2+4x+4-\left(x+2\right)\left(3-x\right)+x^2-6x+9\right)=-18\)
\(\Leftrightarrow x^2+4x+4+x^2-x-6+x^2-6x+9=-\dfrac{18}{5}\)
\(\Leftrightarrow3x^2-3x+\dfrac{53}{5}=0\)
\(\Leftrightarrow x^2-x+\dfrac{53}{15}=x^2-x+\dfrac{1}{4}+\dfrac{197}{60}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{197}{60}=0\)
Vậy phương trình vô nghiệm .
`a)(x+2)^3+(3-x)^3+18=0`
`<=>x^2+6x^2+12x+8+27-27x+9x^2-x^3+18=0`
`<=>15x^2-12x+57=0`
`<=>5x^2-4x+19=0`
`<=>4x^2-4x+1+x^2+18=0`
`<=>x^2+(2x-1)^2=-18` vô lý
=>pt vô nghiệm
