Question

    Bài 1: Cho \(P=\)\(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{z+y}\)

      Tính giá trị của biểu thức P biết rằng: \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)

 

Answer 1

+ TH1 : \(x+y+z+t=0\)

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

\(\Rightarrow P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

+ TH2 : \(x+y+z+t\ne0\)

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

( do \(x+y+z+t\ne0\))

         \(3x=y+z+t\Rightarrow4x=x+y+z+t\)

\(\Rightarrow\)\(3y=z+t+x\Rightarrow4y=x+y+z+t\)

          \(3z=t+x+y\Rightarrow4z=x+y+z+t\)

           \(3t=x+y+z\Rightarrow4t=x+y+z+t\)

\(\Rightarrow4x=4y=4z=4t\Rightarrow x=y=z=t\)

\(\Rightarrow P=4\)