Question

[๖ACE✪Hoàngミ★Việtツ]

Gọi (x₀:y₀) là nghiệm nguyên pt

x³+y³+1=3xy

Tính P=\(\left(1+\frac{x_0}{y_0}\right)\left(1+y_0\right)\left(1+\frac{1}{x_0}\right)\)

giúp mình với

Answer 1

[Đinh Đức Hùng]

\(x^3+y^3+1=3xy\)

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

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

\(\Leftrightarrow\left(x+y+1\right)\left[\left(x+y\right)^2-\left(x+y\right)+1\right]=3xy\left(1+x+y\right)\)

\(\left(x+y+1\right)\left(x^2+y^2+2xy-x-y+1\right)-3xy\left(1+x+y\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)=0\)

Với \(x+y+1\ne0\) thì \(x^2+y^2-xy-x-y+1=0\)

\(\Leftrightarrow x^2+y^2-xy-x-y+1=0\)

\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y+2=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=0\Rightarrow x=y=1\)(thỏa mãn \(x+y+1\ne0\))

\(\Rightarrow P=\left(1+\frac{x_0}{y_0}\right)\left(1+y_0\right)\left(1+\frac{1}{x_0}\right)=\left(1+\frac{1}{1}\right)\left(1+1\right)\left(1+\frac{1}{1}\right)=8\)

Answer 2

[Phan Nghĩa]

Trần Hoàng Việt  thế này có đúng ko ạ? 

\(\hept{\begin{cases}x=3\\y=3\end{cases}\Rightarrow}3=a.1\Rightarrow a=3\)

\(Px_o,y_o\in y=3x\Rightarrow y_o=3.x_o\)

\(P=\frac{x_o+1}{3x_o+1}=\frac{x_o+1}{3"x_o+1"}\)

\(\hept{\begin{cases}x_o=-1\Rightarrow P=kXD\\x_o\ne-1\Rightarrow P=\frac{1}{3}\end{cases}}\)

P/s: Ko chắc :D