\(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

\(=\frac{x^3+y^3+z^3-3xyz}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2}=\frac{\left(x+y+z\right).\left(x^2+y^2+z^2-xy-yz-zx\right)}{2.\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{x+y+z}{2}\)

p/s: áp dụng 7 hàng đẳng thức là làm đc =)

sai đề rồi nhé , đề phải là :

\(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\frac{\left(x-y\right)^3+3xy.\left(x-y\right)+z^3+3xyz}{x^2+2xy+y^2+y^2+2yz+z^2+z^2-2xz+x^2}\)

\(=\frac{\left(x-y+z\right).\left[\left(x-y\right)^2-\left(x-y\right).z+z^2\right]+3xy.\left(x-y+z\right)}{2x^2+2y^2+2z^2+2xy+2yz-2xz}\)

\(=\frac{\left(x-y+z\right).\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{2.\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{\left(x-y+z\right).\left(x^2+y^2+z^2+xy+yz-xz\right)}{2.\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{x-y+z}{2}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)

Đợi anh chút

\(\dfrac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}=\dfrac{y\left(x^2+2xy+y^2\right)}{2x^2+2xy-xy-y^2}=\dfrac{y\left(x+y\right)^2}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\dfrac{y\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{y\left(x+y\right)}{2x-y}\)

\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)

Chúc bn học tốt!

\(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}\)

ĐKXĐ : \(x+y\ne0\Leftrightarrow x\ne-y\)

\(=\frac{5\cdot3x\cdot\left(x+y\right)^2\left(x+y\right)}{5\cdot y\cdot\left(x+y\right)^2}\)

\(=\frac{3x\left(x+y\right)}{y}\)

\(\frac{-6xy\left(x+y\right)^2}{8x^3y\left(x+y\right)}=\frac{-3\left(x+y\right)}{4x^2}\)

\(2,=x^2-3^2=\left(x-3\right)\left(x+3\right)\\ 3,=\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x=4xy\)

x^2-9=x^2-3^2=(x+3)(x-3)
(x+y)^2-(x-y)^2=(x+y+x-y)(x+y-x-y)=2x

x²-3²=(x-3)(x+3)

(x+y)²-(x-y)²=(x+y-x+y)(x+y+x-y)=2y⋅2x=4xy

Ta có: \(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}\)

= \(\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)

=\(-\frac{x+y}{\left(x-y\right)^2}\)

=\(-\frac{x+y}{x^2-2xy+y^2}\)

Phạm Quốc Cường làm đúng rồi đó

k mình nha

thanks