Question

Tìm x, y, z biết

a) \(\dfrac{x}{y+z+1}\) =\(\dfrac{y}{x+y+2}=\dfrac{z}{x+y-3}\)

b)\(6\left(x-\dfrac{1}{y}\right)=3\left(y-\dfrac{1}{2}\right)=2\left(z-\dfrac{1}{x}\right)=xyz-\dfrac{1}{xyz}\)

Giúp mik nha!

Answer 1

a)Ta có: \(\frac{x}{y+z+1}=\frac{y}{x+y+2}=\frac{z}{x+y-3}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

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

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

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

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