xin loi bn nha

mk moi chi hok lop 6 

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chuc bn hok tot nha

2/\(5ay-3bx+ax-15by.\)

\(=\left(5ay-15by\right)-\left(3bx-ax\right)\)

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

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

3/\(x^2+xy+x+1\)

\(=xy+x^2+x+1\)

4/\(x^2y+xy^2-x-y\)

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

=.= mik ko chắc

\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\)

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

\(10x^2+10xy+5x+5y=10x\left(x+y\right)+5\left(x+y\right)=5\left(2x+1\right)\left(x+y\right)\) \(5ay-3bx+ax-15by=a\left(5y+x\right)-3b\left(5y+x\right)=\left(a-3b\right)\left(5y+x\right)\) \(x^3+x^2-x-1=x^2\left(x+1\right)-\left(x+1\right)=\left(x^2-1\right)\left(x+1\right)=\left(x+1\right)^2\left(x-1\right)\) \(2bx-3ay-6by+ax=x\left(2b+a\right)-3y\left(2b+a\right)=\left(x-3y\right)\left(2b+a\right)\)

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

\(10x^2+10xy+5x+5y\)

\(=\left(10x^2+10xy\right)+\left(5x+5y\right)\)

\(=10x\left(x+y\right)+5\left(x+y\right)\)

\(=\left(10x+5\right)\left(x+y\right)\)

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

1:

a) \(4y\left(x-2\right)-5\left(2-x\right)\)

\(=4y\left(x-2\right)+5\left(x-2\right)\)

\(=\left(x-2\right)\left(4y+5\right)\)

b) \(x^2-4x+4-1\)

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

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

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

c) \(5ay-3bx+ax-15by\)

\(=\left(5ay-15by\right)-\left(3bx-ax\right)\)

\(=5y\left(a-3b\right)-x\left(3b-a\right)\)

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

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

2:

a) \(x.\left(x-2\right)+x-2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)

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

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

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}=-3\\x=2\end{matrix}\right.\)

a)10x${}^{}$+10xy+5x+5y

= 10x( x + y ) + 5( x + y )= ( 10x + 5 )( x + y )= 5( 2x + 1 )( x + y )

b)5ay-3bx+ax-15by

= ( 5ay + ax ) - ( 3bx + 15by ) = a( 5y + x ) - 3b( x + 5y ) = ( a - 3b ) ( x + 5y )

a) 5ay - 3bx + ax - 15by

= (5ay + ax) - (3bx + 15by)

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

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

b) x^3 + x^2 - x - 1

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

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

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

c) (2a + b)^2 - (2b + a)^2

= 4a^2 + 4ab + b^2 - 4b^2 - 4ab - a^2

= 3a^2 - 3b^2

= 3 (a^2 - b^2)

d) (8a^3 - 27b^3) - 2a (4a^2 - 9b^2)

= 8a^3 - 27b^3 - 8a^3 + 18ab^2

= 27b^3 + 18ab^2

= 9b^2 (3b + 2a)

\(=\left(a+b\right)\left(x+1\right)\)

\(b,=a\left(x+y\right)+2\left(x+y\right)=\left(a+2\right)\left(x+y\right)\\ c,=x\left(x+y\right)-2\left(x+y\right)=\left(x-2\right)\left(x+y\right)\\ d,=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(2x-y\right)\)

Bài 4

c) x(x - 2) + (x - 2)²

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

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

= 2(x - 2)(x - 1)

d) 2x(x - y)² - 5(y - x)

= 2x(x - y)² + 5(x - y)

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

Bài 5

a) x² - 6x - 2xy + 12y

= (x² - 6x) - (2xy - 12y)

= x(x - 6) - y(x - 6)

= (x - 6)(x - y)

b) 10ax - 5ay - 2x + y

= (10ax - 5ay) - (2x - y)

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

= (2x - y)(5a - 1)

c) x⁴ + x³y - x - y

= (x⁴ + x³y) - (x + y)

= x³(x + y) - (x + y)

= (x + y)(x³ - 1)

= (x + y)(x - 1)(x² + x + 1)

d) x³ + 2x² - 4x - 8

= (x³ + 2x²) - (4x + 8)

= x²(x + 2) - 4(x + 2)

= (x + 2)(x² - 4)

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

= (x + 2)²(x - 2)

e) xy - 5x - y² + 5y

= (xy - 5x) - (y² - 5y)

= x(y - 5) - y(y - 5)

= (y - 5)(x - y)

f) ax - bx - 2cx - 2a + 2b + 4c

= (ax - bx - 2cx) - (2a - 2b - 4c)

= x(a - b - 2c) - 2(a - b - 2c)

= (a - b - 2c)(x - 2)

g) 5x²y + 5xy² - b²x - b²y

= (5x²y + 5xy²) - (b²x + b²y)

= 5xy(x + y) - b²(x + y)

= (x + y)(5xy - b²)

h) 4x³ - 4x² - 9x + 9

= (4x³ - 4x²) - (9x - 9)

= 4x²(x - 1) - 9(x - 1)

= (x - 1)(4x² - 9)

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