Theo AM - GM :

\(a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow a+b+2ab\le\dfrac{a^2+b^2+2ab+2a+2b}{2}\)

\(\Leftrightarrow a^2+b^2+2ab+2a+2b\ge24\)

\(\Leftrightarrow\left(a+b+6\right)\left(a+b-4\right)\ge0\)

\(\Leftrightarrow A\ge4\) (do a,b dương)

Dấu"=" \(\Leftrightarrow a=b=2\)

Ta có: \(A+B+C=0\)

\(\Leftrightarrow3x^2y+5xy^2-2xy+1+2x^2y-7xy^2+6xy-8-5x^2y+4xy^2-4xy+12=0\)

\(\Leftrightarrow2xy^2+5=0\)

\(\Leftrightarrow2x\cdot\left(-2\right)^2+5=0\)

\(\Leftrightarrow8x+5=0\)

\(\Leftrightarrow8x=-5\)

hay \(x=-\dfrac{5}{8}\)

Vậy: \(x=-\dfrac{5}{8}\)

=>5x-8y=0 và x-2y=12

=>5x-8y=0 và 5x-10y=60

=>2y=-60 và 5x=8y

=>y=-30 và 5x=-240

=>x=-48; y=-30

\(\Leftrightarrow3x^2+5x-12=0\)

=>(x+3)(3x-4)=0

=>x=-3 hoặc x=4/3

3x²+5x=12
3x²+5x-12=0
3x²-4x+9x-12=0
x(3x-4)+3(3x-4)=0
(x+3)(3x-4)=0
\(\Rightarrow\)x+3=0 hoặc 3x-4=0
\(\Rightarrow\)x=-3 hoặc x=\(\dfrac{4}{3}\)

\(2x^2+5x=12\)

\(\Leftrightarrow2x^2+5x-12=0\)

\(\Leftrightarrow2x^2+8x-3x-12=0\)

\(\Leftrightarrow2x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+4=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{3}{2}\end{array}\right.\)

Vậy ............

f(-1)=2 và f(1)=12

=>1-a+b=2 và 1+a+b=12

=>-a+b=1 và a+b=11

=>b=6 a=5

=>f(x)=x^2+5x+6

f(x)=0

=>(x+2)(x+3)=0

=>x=-2 hoặc x=-3

\(2x^2+5x=12\)

\(\Leftrightarrow2x^2+5x-12=0\)

\(\Leftrightarrow2x^2+8x-3x-12=0\) .

\(\Leftrightarrow2x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\2x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)