e) \(x^3+1=x\left(x+1\right)\)
\(\Leftrightarrow x^3-x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
f) \(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=4\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{x}+\dfrac{1}{x^2}\right)+\left(y^2-2.y.\dfrac{1}{y}+\dfrac{1}{y^2}\right)+2+2=4\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2=0\\\left(y-\dfrac{1}{y}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\end{matrix}\right.\)
\(\Leftrightarrow x^2=y^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\x=y=-1\\x=1;y=-1\\x=-1;y=1\end{matrix}\right.\)
e: =>(x+1)(x^2-x+1)-x(x+1)=0
=>(x+1)(x^2-2x+1)=0
=>(x-1)^2*(x+1)=0
=>x=1 hoặc x=-1
f: x^2+1/x^2>=2*căn (x^2*1/x^2)=2
y^2+1/y^2>=2*căn (y^2*1/y^2)=2
=>x^2+1/x^2+y^2+1/y^2>=4
Dấu = xảy ra khi x=1/x và y=1/y
=>\(\left\{{}\begin{matrix}x\in\left\{1;-1\right\}\\y\in\left\{1;-1\right\}\end{matrix}\right.\)
