x^4 + 4x^3+ 6x^2+ 4x = y^2
Hướng dẫn: Ta có: x^4 + 4x^3+ 6x^2+ 4x = y^2
⇔ x^4 +4x^3+6x^2+4x +1- y^2=1
⇔ (x+1)^4 – y^2 = 1
⇔ [(x+1)^2 –y] [(x+1)^2+y]= 1
\(\Leftrightarrow\) \(\hept{\begin{cases}\left(x+1\right)^2-y=1\\\left(x+1\right)^2+y=1\end{cases}}\) hoặc \(\hept{\begin{cases}\left(x+1\right)^2-y=-1\\\left(x+1\right)^2+y=-1\end{cases}}\)
\(\orbr{\begin{cases}1-y=1+y\\-1-y=-1+y\end{cases}}\)
⇒ y = 0 ⇒ (x+1)^2 = 1
⇔ x+1 = ±1 ⇒ x = 0 hoặc x = -2
Vậy ( x, y ) = ( 0, 0 ); ( – 2, 0 )
Chúc bạn hk tốt!!!
