8-9x²+1=9-9x²=9(1-x²)

\(x^2\ge0\Leftrightarrow1-x^2\le1\Leftrightarrow9\left(1-x^2\right)\le9\)

vậy giá trị lớn nhất là 9

a) \(A=x-x^2-10=-\left(x^2-x+\frac{1}{4}\right)-\frac{39}{4}=-\left(x-\frac{1}{2}\right)^2-\frac{39}{4}\le-\frac{39}{4}\)với mọi \(x\).

b) \(B=-x^2-2y^2+2xy-2x+10y-40\)

\(=-x^2-y^2-1+2xy-2x+2y-y^2+8y-16-24\)

\(=-\left(x-y+1\right)^2-\left(y-4\right)^2-24\le-24\)với mọi \(x,y\).

a) \(D=\frac{1}{x^2}+\frac{1}{x}+1=\frac{1}{x^2}+2.\frac{1}{x}.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(\frac{1}{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

b) \(E=x^2+y^2+z^2-xy-yz-zx+2021\)

\(=\frac{1}{2}\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)+2021\)

\(=\frac{1}{2}\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]+2021\)

\(=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+2021>0\)

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mn trả lời giúp mik nhé cần gấp

\(D=x^2+y^2+x-6y+5\)

\(D=\left(x^2+x+\frac{1}{2}^2\right)+\left(y^2-6y+9\right)-\frac{17}{4}\)

\(D=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\le-\frac{17}{4}\)

dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)

\(< =>MIN:D=-\frac{17}{4}\)

Trả lời:

\(D=x^2+y^2+x-6y+5=x^2+y^2+x-6y+\frac{1}{4}+9-\frac{17}{4}\)

\(=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-6y+9\right)-\frac{17}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\ge-\frac{17}{4}\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}}\)

Vậy GTNN của D = - 17/4 khi x = -1/2; y = 3