Question

Tìm x, y, z biết:
\(\frac{1}{x-1}=\frac{2}{y-2}=\frac{3}{z-3}\) và \(x+2y+3z=56\)

Answer 1

Aps dụng tính chất dãy tỉ số bằng nhau Ta có:

\(\frac{1+2+3}{x-1+y-2+z-3}=\frac{1+2+3}{x+y+z-1-2-3}=\frac{1+4+9}{x+2y+3z-\left(-4\right)}=\frac{ }{ }\)

=\(\frac{14}{56+4}=\frac{14}{60}=\frac{7}{30}\)

\(\Rightarrow\)\(\frac{1}{x-1}=\frac{7}{30}\)\(\Rightarrow\)x-1=\(\frac{30}{7}\)\(\Rightarrow\)x=\(\frac{37}{7}\)

\(\Rightarrow\)\(\frac{2}{y-2}=\frac{7}{30}\Rightarrow y-2=\frac{60}{7}\)\(\Rightarrow\)y=\(\frac{74}{7}\)

\(\Rightarrow\)\(\frac{3}{z-3}=\frac{7}{30}\Rightarrow z-3=\frac{90}{7}\)\(\Rightarrow\)x=\(\frac{111}{7}\)

Answer 2

Aps dụng tính chất dãy tỉ số bàng nhau, ta có:

\(\frac{1+2+3}{x-1+2y-2+z-3}=\frac{1+4+9}{x-1+2y-4+3z-9}\)=\(\frac{14}{x+2y+3z-1-2-3}=\frac{14}{56-1-2-3}=\frac{14}{50}=\frac{7}{25}\)

\(\Rightarrow\)\(\frac{1}{x-1}=\frac{7}{25}\Rightarrow x=\frac{32}{7}\)

\(\Rightarrow\)\(\frac{4}{2y-4}=\frac{7}{25}\Rightarrow y=\frac{64}{7}\)

\(\Rightarrow\)\(\frac{9}{3z-9}=\frac{7}{25}\Rightarrow z=\frac{96}{7}\)