Toán Tìm giá trị nhỏ nhất P=a^2+ab+b^2-3a-3b+2023 13/11/2021 By Brielle Tìm giá trị nhỏ nhất P=a^2+ab+b^2-3a-3b+2023
$P=a^2+ab+b^2-3a-3b+2023\\ =\dfrac{1}{2}\left(a^2+2ab+b^2\right)+\left(\dfrac{1}{2}a^2-2.\dfrac{1}{\sqrt{2}}a.\dfrac{3}{\sqrt{2}}+\left(\dfrac{3}{\sqrt{2}}\right)^2\right)+\left(\dfrac{1}{2}b^2-2.\dfrac{1}{\sqrt{2}}b.\dfrac{3}{\sqrt{2}}+\left(\dfrac{3}{\sqrt{2}}\right)^2\right)+2014\\ =\dfrac{1}{2}\left(a+b\right)^2+\left(\dfrac{1}{\sqrt{2}}a-\dfrac{3}{\sqrt{2}}\right)^2+\left(\dfrac{1}{\sqrt{2}}b-\dfrac{3}{\sqrt{2}}\right)^2+2014\ge2014$ Dấu “=” xảy ra $<=>\left\{\begin{array}{l} a-b=0\\ \dfrac{1}{\sqrt{2}}a-\dfrac{3}{\sqrt{2}}=0\\ \dfrac{1}{\sqrt{2}}b-\dfrac{3}{\sqrt{2}}=0\end{array} \right.\\<=>a=b=3 $ Trả lời
$P=a^2+ab+b^2-3a-3b+2023\\ =\dfrac{1}{2}\left(a^2+2ab+b^2\right)+\left(\dfrac{1}{2}a^2-2.\dfrac{1}{\sqrt{2}}a.\dfrac{3}{\sqrt{2}}+\left(\dfrac{3}{\sqrt{2}}\right)^2\right)+\left(\dfrac{1}{2}b^2-2.\dfrac{1}{\sqrt{2}}b.\dfrac{3}{\sqrt{2}}+\left(\dfrac{3}{\sqrt{2}}\right)^2\right)+2014\\ =\dfrac{1}{2}\left(a+b\right)^2+\left(\dfrac{1}{\sqrt{2}}a-\dfrac{3}{\sqrt{2}}\right)^2+\left(\dfrac{1}{\sqrt{2}}b-\dfrac{3}{\sqrt{2}}\right)^2+2014\ge2014$
Dấu “=” xảy ra $<=>\left\{\begin{array}{l} a-b=0\\ \dfrac{1}{\sqrt{2}}a-\dfrac{3}{\sqrt{2}}=0\\ \dfrac{1}{\sqrt{2}}b-\dfrac{3}{\sqrt{2}}=0\end{array} \right.\\<=>a=b=3 $
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