Bài 1 : 

a, \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)

\(=x^2+6x+9+x^2-6x+9+2x^2-18\)

\(=4x^2\)

b, \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9=8\)

Bài 2 : 

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

b, \(3\left(3-x\right)-2x\left(x-3\right)=3\left(3-x\right)+2x\left(3-x\right)=\left(3+2x\right)\left(3-x\right)\)

c, \(3x^2+4x-4=3x^2+6x-2x-4=\left(x+2\right)\left(3x-2\right)\)

Sửa đề : \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(9x+9y-9\right)^2-\left(8x+12y+4\right)^2\)

\(=\left(9x+9y-9-8x-12y-4\right)\left(9x+9y-9+8x+12y+4\right)\)

\(=\left(x-3y-13\right)\left(17x+21y-5\right)\)

Đúng là Tú có khác (:

9( x + y - 1 )² - 4( 2x + 3y + 1 )²

= 3²( x + y - 1 )² - 2²( 2x + 3y + 1 )²

= [ 3( x + y - 3 ) ]² - [ 2( 2x + 3y + 1 ) ]²

= ( 3x + 3y - 3 )² - ( 4x + 6y + 2 )²

= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]

= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )

= ( -x - 3y - 5 )( 7x + 9y - 1 )

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

Xí trước phần b

Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)

\(=\frac{bc}{a^2b+ca^2}+\frac{ca}{b^2c+ab^2}+\frac{ab}{c^2a+bc^2}\)

\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}+\frac{c^2a^2}{ab^2c^2+a^2b^2c}+\frac{a^2b^2}{a^2bc^2+ab^2c^2}\)

\(=\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{bc+ab}+\frac{\left(ab\right)^2}{ca+bc}\)

\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

Cách làm khác của phần b ngắn gọn hơn:)

Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

\(=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)

\(=\frac{\left(\frac{1}{a}\right)^2}{ab+ca}+\frac{\left(\frac{1}{b}\right)^2}{bc+ab}+\frac{\left(\frac{1}{c}\right)^2}{ca+bc}\)

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

8a³ - 36a²b + 54ab² - 27b³ - 8 

= ( 8a³ - 36a²b + 54ab² - 27b³ ) - 8 

= ( 2a - 3b )³ - 2³

= ( 2a - 3b - 2 )[ ( 2a - 3b )² + 2( 2a - 3b ) + 4 ]

= ( 2a - 3b - 2 )( 4a² - 12ab + 9b² + 4a - 6b + 4 )

Giúp tui vs

\(x\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)

mà \(x\ge7\)\(\Rightarrow x\in\left\{10;15;30\right\}\)

Vậy \(x\in\left\{10;15;30\right\}\)