A=2x\(^2\)+4x+6=(\(\sqrt{2}\)x)\(^2\)+2\(\sqrt{2}\)x\(\dfrac{2}{\sqrt{2}}\)+(\(\dfrac{2}{\sqrt{2}}\))\(^2\)+4
=(\(\sqrt{2}\)x+\(\dfrac{2}{\sqrt{2}}\))\(^2\)+4
Do(\(\sqrt{2}x+\dfrac{2}{\sqrt{2}}\))\(^2\)\(\)\(\)>=0
=>\(\left(\sqrt{2}x+\dfrac{2}{\sqrt{2}}\right)^2+4\)>=4
Dấu"=" xảy ra khi \(\left(\sqrt{2}x+\dfrac{2}{\sqrt{2}}\right)^2\)=0
=>\(\sqrt{2}x+\dfrac{2}{\sqrt{2}}\)=0
=>x=-1
Vậy minA=4 khi x=-1
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