Ta có:
\(2x-x^{^2}-2\)
\(=-\left(x^{^2}-2x+2\right)\)
\(=-\left(x^{^2}-2x+1\right)\)
\(=-\left(x^{^2}-2x+1\right)-1\)
\(=-\left(x-1\right)^2-1\)
Do \(-\left(x-1\right)^2\le0\)nên \(-\left(x-1\right)^2-1=2x-x^{^2}-2< 0\)hay biểu thức đề cho luôn âm (đpcm)
\(2x-x^2-2=-\left(x-1\right)^2-1\le-1< 0\forall x\)
Trả lời:
\(2x-x^2-2=-\left(x^2-2x+2\right)=-\left(x^2-2x+1+1\right)=-\left[\left(x-1\right)^2+1\right]\)
\(=-\left(x-1\right)^2-1\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-1\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-1\right)^2-1\le-1< 0\forall x\)
Vậy bt luôn âm với mọi x
