2) \(x^4-x^2+1=0\)(1)

Đặt: t=x², khi đó:

(1)\(\Leftrightarrow t^2-t+1=0\)

\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(2\right)\)

\(\Rightarrow\left(2\right)\) vô nghiệm => (1) vô nghiệm

có : (x-y)² \(\ge0,\forall x,y\)

==>x²-2xy+y² \(\ge\)0 \(\forall x,y\)

==> 2.(x²+y²)\(\ge\)2xy +x²+y² \(\forall x,y\)

==> x²+y² \(\ge\)\(\dfrac{\left(x+y\right)^2}{2}=\dfrac{2^2}{2}=2\) ( do x+y=2) \(\forall x,y\)

lại có (x^2-y²)²\(\ge\)0\(\forall x,y\)

==> x⁴+y⁴-2x²y² \(\ge\)0 \(\forall x,y\)

==> 2.(x⁴+y⁴) \(\ge\)2x²y² + x⁴+y⁴ \(\forall x,y\)

==> x⁴+y⁴ \(\ge\)\(\dfrac{\left(x^2+y^2\right)^2}{2}\ge\dfrac{2^2}{2}=2\)

==> đpcm

dấu ''=,, xảy ra <=> \(\left\{{}\begin{matrix}x+y=2\\x-y=0\\x^2-y^2=0\end{matrix}\right.< =>x=y=1}\)

dấu ''=,, xảy ra <=> x=y=1

\(\dfrac{x^2}{x+2}+\dfrac{4}{x-2}=\dfrac{4}{x^2-4}\)

\(\Leftrightarrow\dfrac{x^2}{x+2}+\dfrac{4}{x-2}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{x^2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2\left(x-2\right)+4\left(x+2\right)=4\)

\(\Leftrightarrow x^3-2x^2+4x+8=4\)

\(\Leftrightarrow x^3-2x^2+4x+8-4=0\)

\(\Leftrightarrow x^3-2x^2+4x+4=0\)

PT vô nghiệm vì không thể tìm được x

Vậy : ....

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