Question

tìm x, y, z biết

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

Answer 1

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

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

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

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

=> \(\frac{x}{y+z+1}=\frac{1}{2}\) => y + z + 1 = 2x => y + z + x + 1 = 3x => 1/2 + 1 = 3x => 3/2 = 3x => x = 1/2

=> \(\frac{y}{x+z+1}=\frac{1}{2}\) => x + z + 1 = 2y => x + y + z + 1 = 3y => 1/2 + 1 = 3y

=> 3/2 = 3y => y = 1/2

=> \(\frac{z}{x+y-2}=\frac{1}{2}\) => x + y - 2 = 2z => x + y + z - 2 = 3z => 1/2 - 2 = 3z => -3/2 = 3z => z = -1/2

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