Question

[❖ミ★𝓣𝓻𝓪𝓷𝓰 (𝓯𝓪𝓷 𝓛𝓝𝓓)★彡『ʈєɑɱ❖๖ۣۜƝƘ☆...]

Anh em bạn bè giúp tôi với

Answer 1

[ღ ☪áø /『ʈєɑɱ❖๖ۣۜƝƘ☆』ღ]

Bạn tham khảo theo link nhé :

https://h.vn/cau-hoi/tim-xyz-biet-y-z-1xx-z-2yx-y-3z1x-y-z.161919684879

# Hok tốt !

Answer 2

[Member lỗi thời :>>]

Ta có :

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

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

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}\)

\(=\frac{y+z+x+z+x+y}{x+y+z}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)\(\Rightarrow\hept{\begin{cases}y+z=\frac{1}{2}-x\\x+z=\frac{1}{2}-y\\x+y=\frac{1}{2}-z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{y+z+1}{x}=\frac{\frac{1}{2}-x+1}{x}\\\frac{x+z+2}{y}=\frac{\frac{1}{2}-y+2}{y}\\\frac{x+y-3}{z}=\frac{\frac{1}{2}-z+3}{z}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{y+z+1}{x}=\frac{\frac{1}{2}-x}{x}\\\frac{x+z+2}{y}=\frac{\frac{3}{2}-y}{y}\\\frac{x+y-3}{z}=\frac{\frac{-5}{2}-z}{z}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{y+z+1}{x}=\frac{x}{2}-1\\\frac{x+z+2}{y}=\frac{3y}{2}-1\\\frac{x+y-3}{z}=\frac{-5z}{2}-1\end{cases}}\)

Mà \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\)

\(\Rightarrow\frac{x}{2}-1=\frac{3y}{2}-1=\frac{-5z}{2}-1\)

\(\Rightarrow\frac{x}{2}=\frac{3y}{2}=\frac{-5z}{2}\)

=> x = 3y = -5z

( Tới đây bạn làm nốt nhé ! Mình mỏi tay rồi )