a: =(x-1)^3

b: =(-2x+1)^3

c: =x^3-3x^2y+3xy^2-y^3

=(x-y)^3

Ta có x 3   +   12 x 2   +   48 x   +   64   =   x 3   +   3 x 2 . 4   +   3 . x . 4 2   +   4 3   =   ( x   +   4 ) 3

Đáp án cần chọn là: A

D= 2( x^2  + 3x) -5

= 2( x^2 +2.3/2 .x +(3/2)^2 -9/4) - 5

= 2( x+3/2)^2 - 9/2 - 5

= 2 ( x+3/2)^2 - 19/2

x³-3x²+3x-1

= (x-1)³

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

\(C=4x^2-4x\)

\(C=4x\left(x-1\right)\)

\(D=2x^2+6x-5\)

\(D=2x^2-4x+10x-5\)

\(D=\left(2x^2-4x\right)+\left(10x-5\right)\)

\(D=2x\left(x-2\right)+5\left(2x-1\right)\)

C=4²-4X

C=4X(X-1)

D=2X²+6X-5

D=2X²-4X+10X-5

D=(2X²-4X)+(10X-5)

D=2X(X-2)+5(2X-1)

8x^3-12x^2+6x-1

=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3

=(2x-1)^3

`= (2x)^3 - 3*4x^2*1 + 3*2x*1-1^3`

`= (2x - 1)^3`

 xác định khi 2 x + 1 3 ≠ 0

Suy ra: 2x + 1 ≠ 0 ⇒ x  ≠  - 1/2

a) \(6x^2y+9+x^4y^2=\left(x^2y+3\right)^2\)

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

c) \(25y^4-10y^2+1=\left(5y^2-1\right)^2\)

\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)