cảm ơn xong chẳng có ai :)))

\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)

Các bạn cố gắng giúp mình nha . Mình xin chân thành cảm ơn 

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

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)

Mình làm phần tiếp của bạn Phạm Tuấn Đạt nha .

b) \(\left(x+6\right)^2-2x\left(x+6\right)+\left(x-6\right)\left(x+6\right)\)

\(=x^2+12x+36-2x^2-12x+x^2-36\)

\(=0\)

2. 

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

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

\(4x^2-4x+1-4x^2+12x-x+3+3=0\)

\(7x+7=0\)

\(x=-1\)

\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)

1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)

2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)

3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)

a) \(\left(x^5+4x^3-6x^2\right):4x^2\)

\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

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

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)