(x²– y² + 6x + 9) : (x + y + 3) = (x² + 6x+ 9) – y² : (x + y + 3)

=(x + 3)² – y² : (x + y + 3) = (x + 3 – y) (x + 3 + y) : (x + y + 3) = (x – y + 3)

còn câu c thì sao bn

a) 3x³ + 6x²y

= 3x².(x + 2y)

b) 2x³ - 6x²

= 2x².(x - 2)

c) 18x² - 20xy

= 2x.(9x - 10y)

d) xy + y² - x - y

= (xy + y²) - (x + y)

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

= (x + y)(y - 1)

e) (x²y² - 8)² - 1

= (x²y² - 8 - 1)(x²y² - 8 + 1)

= (x²y² - 9)(x²y² - 7)

= (xy - 3)(xy + 3)(x²y² - 7)

f) x² - 7x - 8

= x² - 8x + x - 8

= (x² - 8x) + (x - 8)

= x(x - 8) + (x - 8)

= (x - 8)(x + 1)

a: \(3x^3+6x^2y\)

\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)

b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)

c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)

d: \(xy+y^2-x-y\)

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

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

e: \(\left(x^2y^2-8\right)^2-1\)

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

\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)

\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)

f: \(x^2-7x-8\)

\(=x^2-8x+x-8\)

\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)

g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)

\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)

\(=2x\left(2x-y\right)\left(5x-3y\right)\)

h: \(x^2-2x+1-y^2\)

\(=\left(x-1\right)^2-y^2\)

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

i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)

\(=2x\left(x+2\right)-x^2\left(x+2\right)\)
\(=\left(x+2\right)\left(2x-x^2\right)=x\cdot\left(x+2\right)\left(2-x\right)\)

k: \(-x^2+6x-9=-\left(x^2-6x+9\right)\)

\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)

l: \(-2x^2+8xy-8y^2\)

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

\(=-2\left(x-2y\right)^2\)

m: \(3x^2+5x-3y^2-5y\)

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

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

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

g) 10x²(2x - y) + 6xy(y - 2x)

= 10x²(2x - y) - 6xy(2x - y)

= 2x(2x - y)(5x - 3y)

h) x² - 2x + 1 - y²

= (x² - 2x + 1) - y²

= (x - 1)² - y²

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

i) 2x(x + 2) + x² (-x - 2)

= 2x(x + 2) - x²(x + 2)

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

k) -9 + 6x - x²

= -(x² - 6x + 9)

= -(x - 3)²

l) 8xy - 2x² - 8y²

= -2(x² - 4xy + 4y²)

= -2(x - 2y)²

m) 3x² + 5x - 3y² - 5y

= (3x² - 3y²) + (5x - 5y)

= 3(x² - y²) + 5(x - y)

= 3(x - y)(x + y) + 5(x - y)

= (x - y)[3(x + y) + 5]

= (x - y)(3x + 3y + 5)

cái này trong sách í

a.

Bạn hãy làm 1 điều coa thể

ok,tui nhất trí với bn nguyễn trọng bằng

\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)

\(=\left(x+\frac{1}{2}\right)\left(x-1\right)\left(x-\frac{2}{3}\right):\left(x+\frac{1}{2}\right)\)

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

a) Ta có: 6x^3 - 7x² - x+2 = 6x³+3x²-10x²-5x+4x+2

                                      = 3x² ( 2x+1) - 5x(2x+1) + 2(2x+1)

                                      = (2x+1)(3x²-5x+2)

Suy ra: (6x³-7x²-x+2): (2x+1)= 3x²-5x+2

b) Ta có: x² -y²+6x+9= (x²+6x+9) - y²

                               = (x+3)² - y²

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

Suy ra: (x²-y²+6x+9): (x+y+3)= x-y+3

Làm vậy nha pạn :) 

nếu tao biết tao chết liền

2 .tìm x

a , x ( x + 2 ) = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b, x ( x-5 )= 5 -x

<=> x ( x-5 ) + x - 5 = 0

<=> x (x-5) + ( x-5)= 0

<=> (x-5)(x+1 )=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

c) ( x + 1 ) ( 6x² + 2x ) + ( x - 1 ) ( 6x² + 2x ) = 0

\(\Leftrightarrow\) ( 6x² + 2x ) \([\)(x+1)(x-1)\(]\)=0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x\left(3x+1\right)=0\\x^{2^{ }}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3x+1=0\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-1}{3}\\x=1\end{matrix}\right.\)

1 ,a) 2a ( x - y ) - ( y - x ) = 2ax - 2ay - y + x

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

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

b) a² ( x - y ) - ( y - x ) = a²x - a²y - y + x

= x ( a²+ 1 ) - y ( a² +1 )

= ( a²+1 ) - (x-y )

c) x ( x - y ) + y ( y - x ) - 3 ( x - y ) = x ² - xy -+ y ² - xy - 3x + 3y

= x² - 2xy + y² -3x + 3y

= (x-y)² -3 ( x - y )

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