Lời giải:
\(A=\left(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}\right)\left(\frac{1}{y-z}+\frac{1}{z-x}+\frac{1}{x-y}\right)-\frac{x}{(y-z)(z-x)}-\frac{x}{(y-z)(x-y)}-\frac{y}{(z-x)(x-y)}-\frac{y}{(z-x)(y-z)}-\frac{z}{(x-y)(y-z)}-\frac{z}{(x-y)(z-x)}\)

\(=0-\frac{x(x-y)+x(z-x)+y(y-z)+y(x-y)+z(z-x)+z(y-z)}{(x-y)(y-z)(z-x)}\)

\(=0-\frac{x^2+xz+y^2+xy+z^2+zy-(xy+x^2+yz+y^2+zx+z^2)}{(x-y)(y-z)(z-x)}=0-\frac{0}{(x-y)(y-z)(z-x)}=0\)

TH1: \(x+y+z+t\ne0\) 

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)

TH1: \(x+y+z+t=0\) 

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

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)

Tham khảo: https://hoc24.vn/cau-hoi/cho-bieu-thuc-pdfracxyztdfracyztxdfracztxydfractxyz-tinh-gia-tri-bieu-thuc-p-biet-dfracxyztdfracyzt.3023321885549

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Lời giải:
Nếu $x+y+z+t=0$ thì:

$P=\frac{-(z+t)}{z+t}+\frac{-(t+x)}{t+x}+\frac{-(x+y)}{x+y}+\frac{-(y+z)}{y+z}$

$=-1+(-1)+(-1)+(-1)=-4$

Nếu $x+y+z+t\neq 0$ thì áp dụng TCDTSBN:

$\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3}$

$\Rightarrow 3x=y+z+t; 3y=z+t+x; 3z=t+x+y; 3t=x+y+z$

$\Rightarrow x=y=z=t$

$\Rightarrow P=1+1+1+1=4$

Bài này dễ thôi:vv

Theo đề ta có: \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\Leftrightarrow\dfrac{xbc+yac+zab}{abc}=0\Leftrightarrow xbc+yac+zab=0\)

Lại có:\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\Rightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=4\)

=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{bc}{yz}+\dfrac{ca}{xz}\right)=4\)

=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{abz+bcx+cay}{xyz}\right)=4\)

=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2.0=4\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=2\)

Vậy...

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

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

Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

Xét \(x+y+z=0\)

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

\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)

Xét \(x+y+z\ne0\) thì ta có:

\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)

\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)

Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)

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