Giải:
a) \(x^2-2xy+y^2+1>0\)
\(\Leftrightarrow\left(x-y\right)^2+1>0\) (luôn đúng)
Vậy ...
b) Ta có:
\(x\le x^2\)
\(\Leftrightarrow x-x^2\le0\)
\(\Leftrightarrow x-x^2-1\le-1\)
\(\Leftrightarrow x-x^2-1< 0\) (đpcm)
Vậy ...
a) Ta có: \(x^2-2xy+y^2+1=\left(x-y\right)^2+1>0;\forall x,y\)
Vì: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0;\forall x,y\\1>0\end{matrix}\right.\)
b) Ta có: \(x-x^2-1=-\left(x^2-x+1\right)\)
...................................= \(-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
...................................= \(-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\)
...................................= \(-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0,\forall x\)
Vì: \(\left\{{}\begin{matrix}-\left(x-\dfrac{1}{2}\right)^2< 0,\forall x\\-\dfrac{3}{4}< 0\end{matrix}\right.\)
a) x2 − 2xy + y2 + 1 > 0
⇔(x−y)2 + 1 > 0
⇔(x−y)2 + 1 > 0
Vậy x2 − 2xy + y2 + 1 > 0
CM:
a) \(x^2-2xy+y^2+1=\left(x-y\right)^2+1\ge1>0\left(ĐFCM\right)\)
\(Vì:\left(x-y\right)^2\ge0\forall x,y\)
b)Ta có:
\(x-x^2-1=-\left(x^2-x+1\right)=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right)-\dfrac{3}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\) (ĐFCM)
\(Vì:-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\)