Question

[Trần Anh Đức]

Tìm x,y,z biết: \(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)(x,y,z khác 0)

Answer 1

[Lê Anh Tú]

Dùng tính chất tỉ lệ thức:

• x+y+z = 0

\(\frac{x}{\left(y+z+1\right)}=\frac{y}{\left(x+z+1\right)}=\frac{z}{\left(x+y-2\right)}=0\Rightarrow x=y=z=0\) 

Áp dụng tính chất tỉ lệ thức: 

\(x+y+z=\frac{x}{\left(y+z+1\right)}=\frac{y}{\left(x+z+1\right)}=\frac{z}{\left(x+y-2\right)}=\left(\frac{x+y+z}{2x+2y+2z}\right)=\frac{1}{2}\)

=> x+y+z = \(\frac{1}{2}\)

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

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

+) \(z=\frac{1}{2}-\left(x+y\right)=\frac{1}{2}-1=\frac{-1}{2}\)

Answer 2

[I don't need you]

TA CÓ: \(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{z+y+1+x+z+1+x+y-2}=\frac{1.\left(x+y+z\right)}{\left(1+1-2\right)+2x+2y+2z}\)

\(=\frac{1.\left(x+y+z\right)}{0+2.\left(x+y+z\right)}=\frac{1.\left(x+y+z\right)}{2.\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

\(\Rightarrow\frac{x}{z+y+1}=\frac{1}{2}\)\(\Rightarrow2x=z+y+1\)\(\Rightarrow3x=x+z+y+1\)\(\Rightarrow3x=\frac{1}{2}+1\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

\(\frac{y}{x+z+1}=\frac{1}{2}\)\(\Rightarrow2y=x+z+1\Rightarrow3y=y+x+z+1\Rightarrow3y=\frac{1}{2}+1=\frac{3}{2}\Rightarrow y=\frac{1}{2}\)

\(\frac{z}{x+y-2}=\frac{1}{2}\)\(\Rightarrow2z=x+y-2\Rightarrow3z=x+y+z-2\Rightarrow3z=\frac{1}{2}-2=\frac{-3}{2}\Rightarrow z=\frac{-1}{2}\)

VẬY X= 1/2; Y= 1/2 ; Z= -1/2

CHÚC BN HỌC TỐT!!!!!!