Question

Bài 1. tìm x,y

a) \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)

Answer 1

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{1+3y}{12}=\frac{1+5y}{5x}+\frac{1+7y}{4x}=\frac{1+3y+1+5y-1-7y}{12+5x-4x}=\frac{\left(1+1-1\right)+\left(3y+5y-7y\right)}{12+\left(5x-4x\right)}=\frac{3+y}{12+x}=\frac{15+5y}{60+5x}\)

\(=\frac{1+5y}{5x}=\frac{15+5y}{60+5x}=\frac{15+5y-1-5y}{60+5x-5x}=\frac{14}{60}=\frac{7}{30}\)

=>\(\frac{1+3y}{12}=\frac{7}{30}=>1+3y=\frac{7}{30}.12=\frac{14}{5}=>3y=\frac{9}{5}=>y=\frac{9}{5}:3=\frac{3}{5}\)

\(\frac{1+5y}{5x}=\frac{7}{30}=>5x=\left(1+5y\right):\frac{7}{30}=\left(1+5.\frac{3}{5}\right).\frac{30}{7}=4.\frac{70}{7}=\frac{120}{17}=>x=\frac{120}{17}:5=\frac{24}{17}\)

=>\(x=\frac{24}{17},y=\frac{3}{5}\)