\(8+8\left(x-2\right)^3\)

\(=8\left[1+\left(x-2\right)^3\right]\)

\(=8\left(1+x-2\right)\left[1-\left(x-2\right)+\left(x-2\right)^2\right]\)

\(=8\left(x-1\right)\left(1-x+2+x^2-4x+4\right)\)

\(=8\left(x-1\right)\left(x^2-5x+7\right)\)

#\(Toru\)

=8[(x-2)^3+1]

=8(x-2+1)[(x-2)^2-(x-2)+1]

=8(x-1)(x^2-4x+4-x+2+1)

=8(x-1)(x^2-5x+7)

8 + 8(x - 2)³

= 8[1 + (x - 2)³]

= 8(1 + x - 2)[1 - x + 2 + (x - 2)²]

= 8(x - 1)(3 - x + x² - 4x + 4)

= 8(x - 1)(x² - 5x + 7)

Trả lời:

7, 49y² - x² + 6x - 9

= 49y² - ( x² - 6x + 9 )

 = ( 7y )² - ( x - 3 )²

= ( 7y - x + 3 ) ( 7y - x - 3 )

8, sửa đề: 25x² - 4y² - 4y - 1

= 25x² - ( 4y² + 4y + 1 )

= ( 5x )² - ( 2y + 1 )

= ( 5x - 2y - 1 ) ( 5x + 2y + 1 )

9, 4x² - y² + 8y - 16

= 4x² - ( y² - 8y + 16 )

= ( 2x )² - ( y - 4 )²

= ( 2x - y + 4 ) ( 2x + y - 4 )

a, \(49y^2-x^2+6x-9=49y^2-\left(x-3\right)^2=\left(7y-x+3\right)\left(7y+x-3\right)\)

b, đề sai rồi bạn 

c, \(4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)

(2x +y-4)

a: =(căn x+4)(căn x-1)

b: =(căn x-1)(x+căn x+1)

\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)

\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)

\(x-6\sqrt{x-3}+6\text{=}x-3-6\sqrt{x-3}+9\)

\(\text{=}\left(\sqrt{x-3}\right)^2-2.3.\sqrt{x-3}+\left(3\right)^2\)

\(\text{=}\left(\sqrt{x-3}-3\right)^2\)

A =  \(x-6\)\(\sqrt{x-3}\) + 6 (đkxd \(x>3\))

A = (\(x\) - 3) - 2.3.\(\sqrt{x-3}\) + 9 

A = (\(\sqrt{x-3}\))² - 2.3.\(\sqrt{x-3}\) + 3²

A = (\(\sqrt{x-3}\)- 3)²

\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=2+\sqrt{3}+\sqrt{6}+2\sqrt{2}\)

\(=2+\sqrt{3}+\sqrt{2}\left(2+\sqrt{3}\right)=\left(2+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)

\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)\)