Question

[Lê Đoàn Song Tú]

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Giair hộ mình bài 6 với ạ 

Answer 1

[Nguyễn Hoàng Minh]

\(6,\\ a,P=9\left(x^2-2\cdot\dfrac{1}{9}x+\dfrac{1}{81}\right)+\dfrac{26}{9}=9\left(x-\dfrac{1}{9}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\\ P_{min}=\dfrac{26}{9}\Leftrightarrow x-\dfrac{1}{9}=0\Leftrightarrow x=\dfrac{1}{9}\\ b,Q=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\\ Q_{min}=\dfrac{1}{4}\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\\ c,R=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\\ R_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)