Question

Giải các phương trình sau:

a, \(9x^2+y^2=18x+6y-18\)

b, \(y^3=x^3+x^2+x+1\) với nghiệm nguyên

Answer 1

a/ \(9x^2+y^2=18x+6y-18\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

Answer 2

a) \(9x^2+y^2=18x+6y-18\)

\(\Rightarrow9x^2+y^2-18x-6y+9=0\)

\(\Rightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)=0\)

\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2=0\)

Mà \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)

Vậy ....................

Câu b để mik nghĩ  tiếp

Answer 3

b/ Ta có:

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

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

\(\Leftrightarrow2x^2+2x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=1\\y=0\end{cases}}\)