\(\left(-\dfrac{1}{2}xy^2\right)^2.6x^4.\left(-2y^2\right)\)
\(\left(\dfrac{1}{2}x-\dfrac{3}{4}\right)^2+\left|y^2+1\right|=1\)
Vì \(\left|y^2+1\right|>0\forall y\in N\)
Và \(\left(\dfrac{1}{2}x-\dfrac{3}{4}\right)^2\ge0\forall x\)
Dấu "=" xảy ra khi \(\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{2}\Rightarrow y=0\)
\(\Rightarrow BT=\left(\dfrac{-1}{2}.1.0^2\right)^2.6x^4.\left(-2^2\right)^2=0\)
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