\(G=3x^2-5x+3\)
\(=x^2+x^2+x^2-2x-2x-x+1+1+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x^2-2x+1\right)+\left(x^2-2x+1\right)+\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-1\right)^2+\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=2\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có :
\(2\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
=> Biểu thức luôn dương với mọi x
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