Question

\(\frac{3x-2}{3}=\frac{2y+1}{4}=\frac{z-6}{5}=\frac{6x+10y+2y-11}{11x+3}\)

Answer 1

ta có 2TH

TH₁ 6x+10y+2z - 11 = 0

\(\Rightarrow\hept{\begin{cases}3x-2=0\\2y+1=0\\z-6=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{1}{2}\\z=6\end{cases}}}\)

TH2 6x+10y+2z - 11 \(\ne\)0

áp dụng t/c dãy tỉ số bằng nhau ta có

\(\frac{3x-2}{3}=\frac{2y+1}{4}=\frac{z-6}{5}=\frac{6x+10y+2z-11}{6+20+10}\)\(=\frac{6x+10y+2z-11}{36}\)

=> 36 = 11x + 3

=> x = 3

\(\Rightarrow\hept{\begin{cases}\frac{3x-2}{3}=\frac{7}{3}\\\frac{2y+1}{4}=\frac{7}{3}\\\frac{z-6}{5}=\frac{7}{3}\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=\frac{25}{6}\\z=\frac{53}{3}\end{cases}}\)