\(1,\\ a,=x^2+6x+9-x^2-6x=9\\ b,=3x-1+6x-9x^2+x-10=-9x^2+10x-11\\ 2,\\ a,=4xy\left(x^2-2xy+y^2\right)=4xy\left(x-y\right)^2\)

Các bạn ơi giúp mình đi , minh đang cần gấp

\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)

\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)

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

\(=-2x^3-13x^2-x-12\)

a) = x³ + 9x² + 27x + 27 - 9x³ -6x² - x + 8x³ +1 -3x² =54

26x +28 = 54

26x = 54-28 = 26

x = 1

b) = x³ - 9x² + 27x -27 - x³ +27 +6x² + 12x + 6 +3x² = -33

39x +6 = -33

39x = -33-6 = -39

x = -1

\(\sqrt{4-x^2}=\sqrt{x+2}\) (ĐK: \(-2\le x\le2\))

\(\Leftrightarrow4-x^2=x+2\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow x^2+2x-x-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

_______

\(\sqrt{9x^2-4}=2\sqrt{3x-2}\) (ĐK: \(x\ge\dfrac{2}{3}\)) 

\(\Leftrightarrow9x^2-4=4\left(3x-2\right)\)

\(\Leftrightarrow9x^2-4=12x-8\)

\(\Leftrightarrow9x^2-12x+4=0\)

\(\Leftrightarrow\left(3x-2\right)^2=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)

a. (x + 3).(x² - 1)

= x.x² - x.1 + 3.x² - 3.1

= x³ - x + 3x² - 3

= x³ + 3x² - x - 3

b. (3x + 2).(4x - 1)

= 3x.4x - 3x + 2.4x - 2

= 12x² - 3x + 8x - 2

= 12x² + 5x - 2

c. (2x - 3).(3x + 2)

= 2x.3x + 2x.2 - 3.3x - 3.2

= 6x² + 4x - 9x - 6

= 6x² - 5x - 6

d. (12x - 5).(4x + 1)

= 12x.4x + 12x - 5.4x - 5

= 48x² + 12x - 20x - 5

= 48x² - 8x - 5

e. (x - 3).(x² + 3x + 9)

= x.x² + x.3x + x.9 - 3x² - 3.3x - 3.9

= x³ + 3x² + 9x - 3x² - 9x - 27

= x³ - 27 (Đây là dạng HĐT x³ - 3³)