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

bạn tự kết luận nhé ! 

a, \(4x-3=2\left(x-3\right)\Leftrightarrow4x-3=2x-6\)

\(\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)

b, \(5x^2+x=0\Leftrightarrow x\left(5x+1\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)

c, \(\left(3x-5\right)\left(x+7\right)=0\Leftrightarrow x=-7;x=\frac{5}{3}\)

d, \(\frac{2}{x-3}-\frac{3}{x+3}=\frac{7x-1}{x^2-9}\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow\frac{2\left(x+3\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{7x-1}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x+6-3x+9=7x-1\Leftrightarrow-x+15=7x-1\)

\(\Leftrightarrow-8x=-16\Leftrightarrow x=2\)( tmđk )

e, \(\left(12x-1\right)\left(6x-1\right)\left(4x-1\right)\left(3x-1\right)=330\)

\(\Leftrightarrow\left(12x-1\right)\left(12x-2\right)\left(12x-3\right)\left(12x-4\right)=330.24=7920\)

\(\Leftrightarrow\left(12x-1\right)\left(12x-4\right)\left(12x-2\right)\left(12x-3\right)=7920\)

\(\Leftrightarrow\left(144x^2-60x+4\right)\left(144x^2-60x+6\right)=7920\)

Đặt \(144x^2-60x+4=t\)

\(t\left(t+2\right)=7920\Leftrightarrow t^2+2t-7920=0\)

\(\Leftrightarrow\left(t-88\right)\left(t+90\right)=0\Leftrightarrow t=88;t=-90\)

suy ra :TH1 :  \(144x^2-60x+4=88\Leftrightarrow12\left(12x+7\right)\left(x-1\right)=0\Leftrightarrow x=-\frac{7}{12};x=1\)

TH2 : \(144x^2-60x+4=-90\Leftrightarrow144x^2-60x+94=0\)

\(\Leftrightarrow x=\frac{5\pm3\sqrt{39}i}{24}\)

T = x² + 2xy + y² - 2x - 2y - 1 

= (x + y)² - 2(x + y) + 1 - 2

= (x + y - 1)² - 2 \(\ge\)-2 

Dấu "=" xảy ra <=> x + y - 1 = 0

=> x + y = 1

Vậy Min A = -2 <=> x + y = 1

(x + y)² = 27²

<=> x² + y² + 2xy = 729

<=> x² + y² + 2.180 = 729

<=> x² + y² = 369

<=> x² + y² - 2xy = 369 - 2xy

<=> (x - y)² = 9

Mà x < y => x - y = -3