a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

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

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

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

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=a+b+c\)

Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:

\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).

Không mất tính tổng quát, giả sử x + y = 0

\(\Leftrightarrow x=-y\)

\(\Leftrightarrow x^3=-y^3\).

Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).

Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).

Bài 1: 

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

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

Để A=0 thì x+1=0

hay x=-1

b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)

Để B=0 thi (x-2)(x+2)=0

=>x=2 hoặc x=-2

Câu a:

Xét tử số:

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

\(=(x-y)^3+z^3+3xy(x-y+z)\)

\(=(x-y+z)[(x-y)^2-z(x-y)+z^2]+3xy(x-y+z)\)

\(=(x-y+z)(x^2+y^2+z^2-2xy-xz+yz)+3xy(x-y+z)\)

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

Xét mẫu số:

\((x+y)^2+(y+z)^2+(z-x)^2\)

\(x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+x^2\)

\(2(x^2+y^2+z^2+xy+yz-xz)\)

Do đó: \(\frac{x^3-y^3+z^3+3xyz}{(x+y)^2+(y+z)^2+(z-x)^2}=\frac{x-y+z}{2}\)

Câu b:

Xét tử số:

\((x^2-y)(y+1)+x^2y^2-1\)

\(=x^2y+x^2-y^2-y+x^2y^2-1\)

\(=(x^2y-y)+(x^2-1)+(x^2y^2-y^2)\)

\(=y(x^2-1)+(x^2-1)+y^2(x^2-1)=(x^2-1)(y^2+y+1)\)

Xét mẫu số:
\((x^2+y)(y+1)+x^2y^2+1\)

\(=x^2y+x^2+y^2+y+x^2y^2+1\)

\(=(x^2y+y)+(x^2+1)+(x^2y^2+y^2)\)

\(=y(x^2+1)+(x^2+1)+y^2(x^2+1)\)

\(=(x^2+1)(y+1+y^2)\)

Do đó:

\(\frac{(x^2-y)(y+1)+x^2y^2-1}{(x^2+y)(y+1)+x^2y^2+1}=\frac{(x^2-1)(y^2+y+1)}{(x^2+1)(y^2+y+1)}=\frac{x^2-1}{x^2+1}\)