`x^3-6x^2y+12xy^2-8y^3`

`=x^3-3.x^2 .2y +3.x.(2y)^2-(2y)^3`

`=(x-2y)^3`

ĐK: \(x\ne2\).

a) \(P=\frac{x+1}{x-2}=\frac{x-2+3}{x-2}=1+\frac{3}{x-2}\)nguyên mà \(x\)nguyên nên \(x-2\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\)

suy ra \(x\in\left\{-1,1,3,5\right\}\).

Thử lại để \(P\)nguyên dương thì \(x\in\left\{-1,3,5\right\}\).

b) \(-x^2-x+2=0\)

\(\Leftrightarrow-x^2+x-2x+2=0\)

\(\Leftrightarrow\left(-x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\Rightarrow P=\frac{1}{4}\\x=1\Rightarrow P=-2\end{cases}}\)

hello

Ta có : (x + 3)(x + 4) > (x - 2)(x + 9) + 25

<=> x² + 7x + 12 > x² + 7x - 18 + 25

<=> 0x > -5 (đúng \(\forall x\))

Vậy bất phương trình có vô số nghiệm 

Đặt A = \(2x^2-2x+1=2\left(x^2-x+\frac{1}{2}\right)=2\left(x^2-x+\frac{1}{4}+\frac{1}{4}\right)=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

=> Min A = 1/2 

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy Min A = 1/2 <=> x = 1/2 

b) Đặt B = \(x^2-x+5=x^2-x+\frac{1}{4}+\frac{19}{4}=\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\)

=> Min B = 19/4

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy Min  B = 19/4 <=> x =1/2

c) Đặt C = \(3x^2-4x+5=3\left(x^2-\frac{4}{3}x+\frac{5}{3}\right)=3\left(x-\frac{2}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\)

=> Min C = 11/3 

Dấu "=" xảy ra <=> x - 2/3 = 0 <=> x = 2/3

Vậy Min C = 11/3 <=> x = 2/3

d) Đặt D = \(2x^2+3x+5=2\left(x^2+\frac{3}{2}x+\frac{5}{2}\right)=2\left(x+\frac{3}{4}\right)^2+\frac{31}{8}\ge\frac{31}{8}\)

=> Min D = 31/8

Dấu "=" xảy ra <=> x + 3/4 = 0 <=>  x =-3/4

Vậy Min D = 31/8 <=> x = -3/4

\(=9x^3-3x^2-9x^2+6x-1\)1

\(=3x^2\left(3x-1\right)-\left(9x^2-6x+1\right)\)

\(=3x^2\left(3x-1\right)-\left(3x-1\right)^2\)

\(=\left(3x-1\right)\left(3x^2-3x+1\right)\)