`a)(x-1)^2-x(x+2)=-3+4x`
`⇒x^2-2x+1-x^2-2x=-3+4x`
`⇒-4x+1=-3+4x`
`⇒-4x-4x=-3-1`
`⇒-8x=-4`
`⇒x=-4:-8`
`⇒x=1/2`
`b)(x-2)(x^2+2x+4)=19`
`⇒x^3+2x^2+4x-2x^2-4x-8=19`
`⇒x^3-8=19`
`⇒x^3=19+8`
`⇒x^3=27`
`⇒x^3=3^3`
`⇒x=3`
`c)(x-1)^2-(2x+3)^2=0`
`⇒(x-1-2x-3)(x-1+2x+3)=0`
`⇒(-x-4)(3x+2)=0`
`⇒(x+4)(3x+2)=0`
`⇒` \(\left[ \begin{array}{l}x+4=0\\3x+2=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=-4\\x=-\dfrac{2}{3}\end{array} \right.\)
`d)(x-1)^3-x(x-2)^2-(x-2)=0`
`⇒x^3-3x^2+3x-1-x(x^2-4x+4)-x+2=0`
`⇒x^3-3x^2+2x+1-x^3+4x^2-4x=0`
`⇒x^2-2x+1=0`
`⇒(x-1)^2=0`
`⇒x-1=0`
`⇒x=1`
`2,`
`C=x^3-y^3-3xy`
`⇒C=(x-y)(x^2+xy+y^2)-3xy`
Mà `x-y=1`
`⇒C=1.(x^2+xy+y^2)-3xy`
`⇒C=x^2+xy+y^2-3xy`
`⇒C=x^2-2xy+y^2`
`⇒C=(x-y)^2`
`⇒C=1`
Vậy giá trị `C=1` khi `x-y=1`
