Cái này y hệt cái đề mik thi:)

trời dài thế làm lâu phết đó nha hừm làm theo đúng công thức là được :)

b) 24x^2+6x^2y−2x−12yx−3y^2x+y 

tôi làm theo cách tìm tích số

nếu thấy đúng thì tick cho tôi nha

a,

\(=15x^3y^2-9x^3y^3+6x^2y^3\)

b

\(=12x^2-2x-3xy^2+y\)

\(a,15x^3y^2-9x^3y^3+6x^3y^3\\ b,12x^3+6x^2y-2x-6x^2y-3xy^2-y\\ =12x^3-2x-3xy^2-y\\ c,4x^2y^3-1\)

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].

a) Đa thức thương 4x – 11 và đa thức dư 26x – 10.

b) Đa thức thương 2 x 2  – 3x + 5 và đa thức dư 3x + 4.

c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)

\(=x^2+x-10\)

a) \(\left(2x^3-x^2+5x\right):x\)

\(=\dfrac{2x^3-x^2+5x}{x}\)

\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)

\(=2x^2-x+5\)

b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)

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

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

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

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

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

\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)

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

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

d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)

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

\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)

\(=-\left(2x^2-4xy+6y^2\right)\)

\(=-2x^2+4xy-6y^2\)

e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)

\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)

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

\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)

f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)

\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)

\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)

\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

1.

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

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

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)