Question

Tìm x,y,z biết:

\(9^{x-1}=\frac{1}{9}\)

\(x-3,8=-15,3-1,3x\)

\(\frac{x+2}{7}=\frac{y-3}{5}=\frac{z}{3}\)và \(x+y-z=-17\)

Answer 1

\(9^{x-1}=\frac{1}{9}\Rightarrow9^{x-1}=9^{-1}\Rightarrow x-1=-1\Rightarrow x=0\)

\(x-3,8=-15,3-1,3x\)

\(\Rightarrow x+1,3x=-15,3+3,8\)

\(\Rightarrow2,3x=-11,5\)

\(\Rightarrow x=-11,5:2,3=-5\)

\(\frac{x+2}{7}=\frac{y-3}{5}=\frac{z}{3}=\frac{x+2+y-3-z}{7+5-3}=\frac{x+y-z-1}{9}=\frac{-17-1}{9}=-\frac{18}{9}=-2\)(t/c dãy tỉ số = nhau)

\(\Rightarrow\frac{x+2}{7}=-2\Rightarrow x+2=-2.7=-14\Rightarrow x=-14-2=-16\)

\(\Rightarrow\frac{y-3}{5}=-2\Rightarrow y-3=-2.5=-10\Rightarrow y=-10+3=-7\)

\(\Rightarrow\frac{z}{3}=-2\Rightarrow z=-2.3=-6\)