a. \(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1\)
Ta có: \(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+1\ge1>0\forall x\)
Vậy ....
b. \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2 +\dfrac{3}{4}>0\forall x\)
Vậy ...
c. \(2x^2+2x+1=x^2+2x+1+x^2=\left(x+1\right)^2+x^2\)
Vì \(\left(x+1\right)^2\ge0\forall x;x^2\ge0\forall x\Rightarrow\left(x+1\right)^2+x^2\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x+1\right)^2=0\\x^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vì x không thể cùng lúc có hai giá trị nên đẳng thức không xảy ra.
\(\Rightarrow\left(x+1\right)^2+x^2>0\forall x\)
Vậy ....
a)\(9x^2-6x+2=\left(3x-1\right)^2+1\)
Với mọi x thì \(\left(3x-1\right)^2>=0\)
=>\(\left(3x-1\right)^2+1>=1>0\)
=>\(9x^2-6x+2\)luôn dương
b)\(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Với mọi x thì \(\left(x+\dfrac{1}{2}\right)^2>=0\)
=>\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\)
=>....(đpcm)
c)\(2x^2+2x+1=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)
Với mọi x thì \(2\left(x+\dfrac{1}{2}\right)^2>=0\)
=>\(2\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}>=\dfrac{1}{2}>0\)
=>\(2x^2+2x+1>0\)(đpcm)
