Em tham khảo:

cho 3 số x,y,z đôi một khác nhau và x+y+z=0 Tính\(P=\dfrac{2018\left(x-y\right)\left(y-z\right)\left(z-x\right)}{2xy^2+2... - Hoc24

.

a.\(\left(x^2-y^2-z^2\right)=\left(x-y\right)^2-2z\left(x-y\right)+z^2=x^2-2xy+y^2-2zx+2zy+z^2\)

b.\(\left(x+y-z\right)^2=\left(x+y\right)^2-2z\left(x+y\right)+z^2=x^2+2xy+y^2-2zy-2zx+z^2\)

Ta có \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^3=-z^3\)

\(\Leftrightarrow x^3+y^3+z^3+3xy\left(x+y\right)=0\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow x^3+y^3+z^3=3xyz\)

Đặt \(A=2xy^2+2yz^2+2zx^2+3xyz=2xy^2+2yz^2+2zx^2+x^3+y^3+z^3\)

\(=x^2\left(2z+x\right)+y^2\left(2x+y\right)+z^2\left(2y+z\right)\)

Do \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}2z+x=z-y\\2x+y=x-z\\2y+z=y-x\end{matrix}\right.\)

\(\)\(\Rightarrow A=x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\)

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

\(=\left(x^2-y^2\right)\left(z-y\right)-\left(z^2-y^2\right)\left(x-y\right)\)

\(=\left(x-y\right)\left(z-y\right)\left(x+y-z-y\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

\(\Rightarrow\dfrac{2018\left(x-y\right)\left(y-z\right)\left(x-z\right)}{A}=2018\)

\(\Rightarrow P=2018\)

Vậy \(P=2018\)

\(\left(x+y+z\right)^2\)

\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2\)

\(=x^2+2xy+y^2+2xz+2xy+z^2\)

a, \(\left(x+y+z\right)^2=\left(x+y\right)^2+2\left(x+y\right)z+z^2\)\(=x^2+2xy+y^2+2zx+2zy+z^2=x^2+y^2+z^2+2xy+2yz+2zx\)(đpcm)

b, \(\left(x+y+z\right)^3=\left(\left(x+y\right)+z\right)^3=\left(x+y\right)^3+z^3+3\left(x+y\right)z\left(x+y+z\right)\)

\(=x^3+y^3+3xy\left(x+y\right)+z^3+3\left(x+y\right)z\left(x+y+z\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+z\left(x+y+z\right)\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+zx+zy+z^2\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(y\left(x+z\right)+z\left(x+z\right)\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)