a)Ta có:
\(a+b+ab=a^2+b^2\).
\(\Leftrightarrow a^2-ab+b^2=a+b\).
Ta có:
\(P=a^3+b^3+2020\).
\(P=\left(a+b\right)\left(a^2-ab+b^2\right)+2020\).
\(P=\left(a+b\right)\left(a+b\right)+2020\)(vì \(a^2-ab+b^2=a+b\)).
\(P=\left(a+b\right)^2+2020\).
Ta có:
\(\left(a+b\right)^2\ge0\forall a;b\).
\(\Rightarrow\left(a+b\right)^2+2020\ge2020\forall a;b\).
\(\Rightarrow P\ge2020\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}a+b+ab=a^2+b^2\\\left(a+b\right)^2=0\end{cases}}\Leftrightarrow a=b=0\).
Vậy \(maxP=2020\Leftrightarrow a=b=0\).
b)\(A=\frac{27-12x}{x^2+9}\).
Vì \(x^2+9>0\forall x\)nên \(A\)luôn được xác định.
\(A=\frac{27-12x}{x^2+9}=\frac{4x^2-4x^2+27-12x}{x^2+9}=\frac{\left(4x^2+36\right)-\left(4x^2+12x+9\right)}{x^2+9}\)
\(A=\frac{4\left(x^2+9\right)-\left(2x+3\right)^2}{x^2+9}=4-\frac{\left(2x+3\right)^2}{x^2+9}\).
Ta có:
\(\left(2x+3\right)^2\ge0\forall x\).
\(\Rightarrow\frac{\left(2x+3\right)^2}{x^2+9}\ge0\forall x\)(vì \(x^2+9>0\forall x\)).
\(\Rightarrow-\frac{\left(2x+3\right)^2}{x^2+9}\le0\forall x\).
\(\Rightarrow4-\frac{\left(2x+3\right)^2}{x^2+9}\le4\forall x\).
\(\Rightarrow A\le4\).
Dấu bằng xảy ra.
\(\Leftrightarrow\left(2x+3\right)^2=0\Leftrightarrow x=-\frac{3}{2}\).
Vậy \(maxA=4\Leftrightarrow x=-\frac{3}{2}\).
