Question

1)tìm x;y:

a)\(\frac{x-y}{x+2y}=\frac{3}{4}\)              b)\(\frac{3x-7}{8}=\frac{5}{2}\)            

2)

tìm x;y;z biết : \(\frac{x}{3}=\frac{y}{2}=\frac{z}{6}\)và\(5.x^2+y^2-z^2=117\)

Answer 1

Ta có:

\(\frac{x-y}{x+2y}=\frac{3}{4}\)

\(\Rightarrow\left(x-y\right).4=\left(x+2y\right).3\)

\(\Rightarrow4x-4y=3x+6y\)

\(\Rightarrow4x=3x+10y\)

\(\Rightarrow x=10y\)

Thay \(x=10y\) vào \(\frac{x-y}{x+2y}=\frac{3}{4}\), ta có:

\(\frac{10y-y}{10y+2y}=\frac{3}{4}\)

\(\Rightarrow\frac{9y}{12y}=\frac{3}{4}\)

êk? thôi chắc chịu, pai pai, cứ để hiện lên cho oách

Answer 2

2, Đặt \(\frac{x}{3}=\frac{y}{2}=\frac{z}{6}=a\)

\(\Rightarrow x=3a;y=2a;z=6a\)

\(5x^2+y^2-z^2=117\Rightarrow5.\left(3a\right)^2+\left(2a\right)^2-\left(6a\right)^2=117\)

\(\Rightarrow13a^2=117\Rightarrow a^2=9\)\(\Rightarrow a=3\) hoặc \(a=-3\)

+ Với \(a=3\) thì \(x=3.3=9;y=3.2=6;z=3.6=18\)

+Với \(a=-3\) thì \(x=-9;y=-6;z=-18\)