\(\frac{1+3y}{12}=\frac{1+6y}{16}\)
<=> (1 + 3y).16 = (1 + 6y).12
<=> 16 + 48y = 12 + 72y
<=> 16 - 12 = 72y - 48y
<=> 24y = 4
<=> y = 1/6
Thay y = 1/6 vào đề bài ta được:
\(\frac{1+3.\frac{1}{6}}{12}=\frac{1+9.\frac{1}{6}}{4x}\)
\(\Leftrightarrow\frac{\frac{3}{2}}{12}=\frac{\frac{5}{2}}{4x}\Leftrightarrow6x=30\Leftrightarrow x=5\)
Vậy x = 5; y = 1/6
tìm x thỏa mãn: \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
Tìm x thỏa mãn:
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
1+3y/12=1+6y/16=1+9y/4x
tìm x , y , z biết
a,
\(\frac{x+y}{x}=\frac{y}{x+z}=\frac{z}{x+y}=x+y+z\)
b,
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
c,
\(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}\)
d,
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
tìm x y 1+3y/12=1+6y/5x=1+9y/4x
Tìm x,y biết :\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
tìm x, y biết
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
Tìm x,y biết: \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
a) \(\frac{x}{3}=\frac{y}{7};xy=84\)
b) \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
