a)Thay x=2(TMDK) vào bt Q :

\(Q=\dfrac{2+1}{2^2-9}=-\dfrac{3}{5}\)

b) \(P=\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\\ =\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\\ =\dfrac{2x^2-1-\left(x-1\right)\left(x+1\right)+3x}{x\left(x+1\right)}\\ =\dfrac{2x^2-1-\left(x^2-1\right)+3x}{x\left(x+1\right)}\\ =\dfrac{x^2+3x}{x\left(x+1\right)}=\dfrac{x\left(x+3\right)}{x\left(x+1\right)}=\dfrac{x+3}{x+1}\)

c) \(M=P.Q=\dfrac{x+3}{x+1}.\dfrac{x+1}{x^2-9}\\ =\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x-3}\)

\(M=-\dfrac{1}{2}\\ =>\dfrac{1}{x-3}=-\dfrac{1}{2}\\ =>x-3=-2\\ =>x=1\left(TMDK\right)\)

a) \(5\left(x+2y\right)-15x\left(x+2y\right)=\left(x+2y\right)\left(5-15x\right)\\ =5\left(x+2y\right)\left(1-3x\right)\)

b) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2\\=\left(2x-3\right)^2\)

c) \(\left(3x-2\right)^3-3\left(x-4\right)\left(x+4\right)+\left(x-3\right)^3-\left(x+1\right)\left(x^2-x+1\right)\\ =27x^3-54x^2+18x-8-3\left(x^2-16\right)+x^3-9x^2+27x-27-\left(x^3+1\right)\\=27x^3-54x^2+18x-8-3x^2+48+x^3-9x^2+27x-27-x^3-1\\ =27x^3-57x^2+36x+12\\ =3\left(3x^3-19x^2+12x+4\right)\)

c) \(27x^3-54x^2+36x-8-3x^2+48+x^3-9x^2+27x-27-x^3-1\\ =27x^3-66x^2+63x+12\\=3\left(9x^3-22x^2+21x+4\right)\)

Theo hằng đẳng thức đáng nhớ:

Lời giải:

$A=(x^2-2xy+y^2)+y^2+2x-6y+2028$

$=(x-y)^2+2(x-y)+(y^2-4y)+2028$

$=(x-y)^2+2(x-y)+1+(y^2-4y+4)+2023$

$=(x-y+1)^2+(y-2)^2+2023\geq 0+0+2023=2023$
Vậy $A_{\min}=2023$.

Giá trị này đạt tại $x-y+1=y-2=0$

$\Leftrightarrow y=2; x=1$

$A=(x^2-2xy+y^2)+y^2+2x-6y+2028$

$=(x-y{)}^2+2(x-y)+(y^2-4y)+2028$

$=(x-y{)}^2+2(x-y)+1+(y^2-4y+4)+2023$

$=(x-y+1{)}^2+(y-2{)}^2+2023\ge 0+0+2023=2023$
Vậy $A^{\min }=2023$.

Giá trị này đạt tại $x-y+1=y-2=0$

$\iff y=2;x=1$