Lời giải:
q.
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(2x+5)(x-3)^2}{(3x-1)(x-3)^2}=\frac{2x+5}{3x-1}\)
u.
\(\frac{x^3-y^3+z^3+3xyz}{(x+y)^2+(y+z)^2+(z-x)^2}=\frac{(x+z)^3-3xz(x+z)-y^3+3xyz}{2(x^2+y^2+z^2+xy+yz-xz)}\)
\(=\frac{(x+z)^3-y^3-3xz(x+z-y)}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{(x+z-y)[(x+z)^2+(x+z)y+y^2]-3xz(x+z-y)}{2(x^2+y^2+z^2+xy+yz-xz)}\)
\(=\frac{(x+z-y)[(x+z)^2+(x+z)y+y^2-3xz]}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{(x+z-y)(x^2+y^2+z^2+xy+yz-xz)}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{x+z-y}{2}\)
ư.
\(\frac{x^3+y^3+z^3-3xyz}{(x-y)^2+(y-z)^2+(z-x)}=\frac{x+y+z}{2}\) vì tương tự như phần u, chỉ cần thay $y\to -y$
q) Ta có: \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\dfrac{2x^3+5x^2-12x^2-30x+18x-45}{3x^3-x^2-18x^2+6x+27x-9}\)
\(=\dfrac{x^2\left(2x+5\right)-6x\left(2x+5\right)+9\left(2x+5\right)}{x^2\left(3x-1\right)-6x\left(3x-1\right)+9\left(3x-1\right)}\)
\(=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)
\(=\dfrac{2x+5}{3x-1}\)
