Bài 1: 

a) \(a^2-6a+9=\left(a-3\right)^2\)

b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)

Bài 2:

a)  \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)

\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)

b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)

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

b) \(x^2+y^2+2xy=\left(x+y\right)^2\)

c) \(9x^2+12x+4=\left(3x+2\right)^2\)

d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)

chỉ cần thu gọn đa thức này thôi

a: \(\left(3x-1\right)\left(9x^2+3x+1\right)=27x^3-1\)

b: \(\left(1-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{5}+1\right)=1-\dfrac{x^3}{125}\)

c: \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

d: \(\left(4x+3y\right)\left(16x^2-12xy+9y^2\right)=64x^3+27y^3\)

\(\dfrac{x^2}{4}-3x+9=\left(\dfrac{x}{2}-3\right)^2\)

Ta có  x 3 + 3 x 2 + 3 x + 1 = x 3 + 3 x 2 . 1 + 3 x . 1 2 + 1 3 = ( x + 1 ) 3 .

Ta có

( x 2   +   3 ) ( x 4   –   3 x 2   +   9 )   =   ( x 2   +   3 ) x 2 2 - 3 x 2 + 3 2 = x 2 3 + 3 3

Đáp án cần chọn là: A

Ta có:  ( x + 1 ) ( x 2 - x + 1 ) = x 3 + 1 3 = x 3 + 1 .