Question

[son]

tim cac ng nguyen cua pt sau

x² = y(y+1).(y+2).(y+3)

Answer 1

[Thắng Nguyễn]

\(x^2=y\left(y+1\right)\left(y+2\right)\left(y+3\right)\)

\(\Leftrightarrow x^2=\left(y^2+3y\right)\left(y^2+3y+2\right)\)

Đặt \(y^2+3y=t\) ta có:

\(VP=t\left(t+2\right)=t^2+2t\)

• Nếu \(t>0\)thì\(t^2< x^2=t^2+2t< \left(t+1\right)^2\) suy ra vô nghiệm
• Nếu \(t< -2\)thì\(2t+4< 0\)nên\(t^2+2t>t^2+4t+4=\left(t+2\right)^2\)

Suy ra \(x^2=t^2+2t>\left(t+2\right)^2\left(1\right)\)

Lại có: \(t^2+2t< t^2\Rightarrow x^2< t^2\left(2\right)\)

Từ (1) và (2) suy ra \(\left(t+2\right)^2< x^2< t^2\Rightarrow x^2=\left(t+1\right)^2\)

\(\Rightarrow t^2+2t=\left(t+1\right)^2\left(=x^2\right)\)

Suy ra \(t^2+2t=t^2+2t+1\)(vô lí)

• Nếu \(t=-1\Rightarrow x^2=t^1+2t=-1< 0\)(vô lí)
• Nếu \(t=0\Rightarrow x=0\Rightarrow\hept{\begin{cases}y=-1\\y=-2\\y=-3\end{cases}};y=0\)

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