Question

Tìm các số x, y, z biết rằng:

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

b) \(\dfrac{x-1}{2}\)= \(\dfrac{y-2}{3}\)= \(\dfrac{z-3}{4}\) và 2x+3y-z=50

Mn giúp mình nha

Answer 1

phần a sai đề ak?

phải là \(\frac{z+y+1}{y} \) chứ

Answer 2

\(\frac{y+z+1}{x} \)

Answer 3

Ta có:\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2.\left(x+y+z\right)}{x+y+z}=\dfrac{1}{x+y+z}\)=> \(2.\left(x+y+z\right)=1\)

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

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

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

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

\(\dfrac{\dfrac{1}{2}+1}{x}=3\)

\(\dfrac{\dfrac{3}{2}}{x}=3\)

\(x=\dfrac{3}{2}:3=\dfrac{1}{2}\)

Tương tự đối với tìm y

Ta sẽ có : \(y=\dfrac{5}{6}\)

=> \(z=x+y+z-x-y=\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{5}{6}=\dfrac{-5}{6}\)

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