a) Ta có P = 4 x 2 ( x − 2 y ) 2 ( x + 2 y ) 2 . ( x + 2 y ) 2 16 x = x 4 ( x − 2 y ) 2  

Với x ≠ 0 ,   x ≠   ± 2 y  

b) Ta có Q = 16 x ( x 2 − 16 ) 2 . x 2 − 16 2 x = 8 16 − x 2  với x ≠ 0 ,    x ≠   ± 4

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

\(x^2y^2-x^2+8x-16.\)

\(=x^2y^2-x^2+4x+4x-16+x^2y-x^2y+4xy-4xy\)

\(=x^2y^2+x^2y-4xy-x^2y-x^2+4x+4xy+4x-16\)

\(=\left(x^2y^2+x^2y-4xy\right)-\left(x^2y+x^2-4x\right)+\left(4xy+4x-16\right)\)

\(=xy\left(xy+x-4\right)-x\left(xy+x-4\right)+4\left(xy+x-4\right)\)

\(=\left(xy+x-4\right)\left(xy-x+4\right)\)

ta có 4 x 3 y 2   –   8 x 2 y 3   =   4 x 2 y 2 . x   –   4 x 2 y 2 . 2 y   =   4 x 2 y 2 ( x   –   2 y )    

Vậy 4x³y² – 8x²y³ = 4x²y²(x – 2y)      

Đáp án cần chọn là: C

bấm đúng cho mik đi 

a: =3(x^2-y^2-4x+4y)

=3[(x-y)(x+y)-4(x-y)]

=3(x-y)(x+y-4)

b: \(=4x\left(x^2+y^2+2xy-16\right)\)

\(=4x\left[\left(x+y\right)^2-16\right]\)

\(=4x\left(x+y+4\right)\left(x+y-4\right)\)

c: \(=\left(x+4\right)^2-y^2=\left(x+4+y\right)\left(x+4-y\right)\)

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

a)  ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014 

Đăngt thức xay ra khi x=y=1

a: \(\Leftrightarrow x^2-2x+1+y^2+4y+4=0\)

=>(x-1)^2+(y+2)^2=0

=>x=1 và y=-2

b: \(\Leftrightarrow2x^2+2y^2-16x+32+16y+32=0\)

\(\Leftrightarrow2\left(y-4\right)^2+2\left(x+4\right)^2=0\)

=>y=4; x=-4