Question

Cho biểu thức \(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}+\dfrac{t+z}{z+y}\)

Tìm giá trị của P, biết rằng \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)

Answer 1

Ta có :

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

\(\Rightarrow\dfrac{x+y+z+t}{x+y+t}=\dfrac{x+y+z+t}{x+z+t}=\dfrac{x+y+z+t}{x+y+z}\)

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

* Nếu \(x+y+z+t=0\)

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

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

Thay vào A ta được: \(P=-1+-1=-2\)

*Nếu \(x+y+z+t\ne0\)

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

Làm tương tự tự ta suy ra được \(x=y=z=t\)

=> \(x+y=z+t\)

\(y+z=t+x\)

Thay vào A ta được A= 1+1=2

Vậy... tik mik nha !!!

Answer 2

Xét:

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

\(\Leftrightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)

+ TH1: Nếu \(x+y+z+t\ne0\Rightarrow x=y=z=t\Rightarrow P=4\)

+ TH2: Nếu \(x+y+z+t=0\Rightarrow P=-4\)

Vậy \(\left[{}\begin{matrix}P=4\\P=-4\end{matrix}\right.\)