Question

Cho \(\frac{x}{-2}=\frac{9}{y}=\frac{3-2z}{5}\) và \(x+z=0\). Khi đó \(y=...\)

Answer 1

Ta có: \(\frac{x}{-2}=\frac{2x}{-4}=\frac{3-2z}{5}=\frac{2x-\left(3-2z\right)}{-4-5}=\frac{2x+2z-3}{-9}=\frac{2\left(x+z\right)-3}{-9}=\frac{2.0-3}{-9}=\frac{-3}{-9}=\frac{1}{3}\)

=>\(\frac{9}{y}=\frac{1}{3}\)

=>y=3.9=27

Vậy y=27

Answer 2

Quên ghi chứ "theo t/c của dãy TSBN"

Answer 3

ta có \(\frac{x}{-2}=\frac{-2x}{4}=\frac{3-2z}{5}=\frac{-2x+3-2z}{4+5}=\frac{3-2\left(x+z\right)}{9}=\frac{3}{9}=\frac{1}{3}\)(vì x +z = 0)

 

từ đó \(\Rightarrow\frac{9}{y}=\frac{1}{3}\Rightarrow y=\frac{9.3}{1}=27\)