Question

1)Tìm x,y,z biết

x/3=y/25 và x+y+z=180,z=4x

Answer 1

Ta có: x+y+z =180

<=>5x+y=180

Ta lại có \(\dfrac{x}{3}=\dfrac{5x}{15}=\dfrac{y}{25}\)

Áp dụng dãy tỉ số bằng nhau:

k=\(\dfrac{x}{3}=\dfrac{5x}{15}=\dfrac{y}{25}\)=\(\dfrac{5x+y}{15+25}=\dfrac{180}{40}=\dfrac{9}{2}\)

=>k=\(\dfrac{9}{2}\)

Do đó: \(\dfrac{x}{3}=\dfrac{9}{2}=>x=\dfrac{27}{2}\)

\(\dfrac{y}{25}=\dfrac{9}{2}=>y=\dfrac{225}{2}\)

z=4x=4*\(\dfrac{27}{2}=54\)

Answer 2

\(z=4x\)

\(\Rightarrow x+y+z=x+4x+y=180\)

\(\Rightarrow5x+y=180\)

\(\dfrac{x}{3}=\dfrac{y}{25}\Rightarrow25x=3y\Rightarrow y=\dfrac{25}{3}x\)

\(\Rightarrow5x+\dfrac{25}{3}x=180\)

\(\Rightarrow\dfrac{40}{3}x=180\)

\(\Rightarrow x=180:\dfrac{40}{3}=\dfrac{27}{2}\)

\(\Rightarrow y=\dfrac{27}{2}:3.25=\dfrac{225}{2}\)

\(\Rightarrow z=\dfrac{27}{2}.4=54\)