Question

tìm x,y

\(\frac{1+3y}{12}\) = \(\frac{1+6y}{16}\) = \(\frac{1+9y}{4x}\)

Answer 1

/hoi-dap/question/151622.html

Answer 2

ta có: \(\frac{1+3y}{12}=\frac{1+6y}{16}\)

\(\Rightarrow\frac{2.\left(1+3y\right)}{24}=\frac{1+6y}{16}\)

\(\Rightarrow\frac{2+6y}{24}=\frac{1+6y}{16}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có

\(\frac{2+6y}{24}=\frac{1+6y}{16}=\frac{2+6y-\left(1+6y\right)}{8}=\frac{2+6y-1+6y}{8}=\frac{1}{8}\)

\(\Rightarrow8.\left(1+6y\right)=16\)

\(\Rightarrow8+48y=16\)

\(\Rightarrow48y=8\)

=> y=\(\frac{1}{8}\)

Ta có

\(\frac{2+6y}{24}=\frac{1+6y}{16}=\frac{1}{8}\)

\(\Rightarrow\frac{1+9y}{4x}=\frac{1}{8}\)

Thế y=1/6 vào biểu thức, ta có

\(\frac{1+9y}{4x}=\frac{1}{8}\)

\(\Rightarrow\frac{1+9.\frac{1}{6}}{4x}=\frac{1}{8}\)

\(\Rightarrow\frac{\frac{5}{2}}{4x}=\frac{1}{8}\)

\(\Rightarrow20=4x\)

\(\Rightarrow x=5\)

Vậy x=5