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 \(maxA=4\Leftrightarrow x=-\frac{3}{2}\).

\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{\left(2y\right)^2+2020}\right)=2020\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y+\sqrt{\left(2y\right)^2+2020}=\sqrt{x^2+2020}-x\\x+\sqrt{x^2+2020}=\sqrt{\left(2y\right)^2+2020}-2y\end{matrix}\right.\)

\(\Rightarrow x+2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}=-x-2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}\)

\(\Leftrightarrow2\left(x+2y\right)=0\)

\(\Leftrightarrow x=-2y\)

\(\Rightarrow B=2y^2-8y^2+3y^2-2y+3y+15\)

\(\Rightarrow B=-3y^2+y+15=-3\left(y-\dfrac{1}{6}\right)^2+\dfrac{181}{12}\)

\(B_{max}=\dfrac{181}{12}\) khi \(y=\dfrac{1}{6}\)

a. Vì \(\left|x-1\right|\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left|x-1\right|+\left(y+2\right)^2\ge0\forall x;y\)

\(\Rightarrow\left|x-1\right|+\left(y+2\right)^2+2020\ge2020\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=0\\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\y=-2\end{cases}}}\)

Vậy Bmin = 2020 <=> x = 1 và y = - 2

b. Vì \(x^2\ge0\forall x\Rightarrow-x^2\le0\)

\(\Rightarrow-x^2+2019\le2019\)

Dấu "=" xảy ra \(\Leftrightarrow-x^2=0\Leftrightarrow x=0\)

Vậy Pmax = 2019 <=> x = 0

Vì \(\left|y-1\right|\ge0\forall y;\left(t+2\right)^4\ge0\forall t\)

\(\Rightarrow-\left|y-1\right|-\left|t+2\right|^4\le0\forall y;t\)

\(\Rightarrow-\left|y-1\right|-\left|t-2\right|^4+21\le21\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|y-1\right|=0\\\left|t+2\right|^4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y-1=0\\t+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}y=1\\t=-2\end{cases}}\)

Vậy Qmax <=> y = 1 và t = 2

Cảm ơn bạn Death Note nha

\(M=x^2-8x+5\)

\(\Leftrightarrow M=x^2-8x+16-11\)

\(\Leftrightarrow M=\left(x-4\right)^2-11\ge-11\)

Min M = -11 

\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)

\(N=-3x-6x-9\)

\(\Leftrightarrow N=-9x-9\le-9\)

Max N = -9

\(\Leftrightarrow x=0\)

a) Ta có : M = x² - 8x + 5 = x² - 8x + 16 - 17 = (x - 4)² - 17 \(\ge\)-17

Dấu "=" xảy ra <=> x - 4 = 0 => x = 4