a.
\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)
Đặt \(x^2-x+1=y\) ta được:
\(y\left(y+1\right)=12\)
\(\Leftrightarrow y^2+y-12=0\)
\(\Leftrightarrow y^2+4y-3y-12=0\)
\(\Leftrightarrow\left(y-3\right)\left(y+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=3\\y=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=3\\x^2-x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b.
\(3y^3-7y^2-7y+3=0\)
\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)
\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(3y-1\right)\left(y-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=\dfrac{1}{3}\\y=3\end{matrix}\right.\)
c.
\(x^2+2y^2=4x+4y-6=0\)
\(\Leftrightarrow x^2-4x+4+2y^2-4y+2=0\)
\(\Leftrightarrow\left(x-2\right)^2+2\left(y-1\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) với mọi x;y
\(\Rightarrow\left(x-2\right)^2+2\left(y-1\right)^2\ge0\) ; \(\forall x;y\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
