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10 tháng 9 2018

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

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

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

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

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

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

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

6 ) \(2x^2-6xy+5x-15y\)

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

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

7 ) \(ax^2-3axy+bx-3by\)

\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)

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

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

8 ) \(x^2+4x-5x-20=0\)

\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

9 ) \(x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

10 ) \(x^2-6x-4x+24=0\)

\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

:D

8 tháng 9 2020

A = 10ax - 5ay - 2x + y

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

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

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

B = 2x2 - 6xy + 5x - 15y 

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

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

C = ax2 - 3axy + bx - 3by

= ( ax2 + bx ) - ( 3axy + 3by )

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

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

D = 2ax3 + 6ax2 + 6ax + 18a

= 2ax2( x + 3 ) + 6a( x + 3 )

= ( x + 3 )( 2ax2 + 6a )

= ( x + 3 )2a( x2 + 3 )

E = 5x2y + 5xy2 + a2x + a2y ( đã sửa 1 dấu '-' )

= 5xy( x + y ) + a2( x + y )

= ( x + y )( 5xy + a2 )

F = 10xy2 - 5by2 + 2a2x - aby ( xem lại đề chứ không phân tích được :)) )

25 tháng 7 2017

Bài 1 : 

a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)

b)  \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)

c)  \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)

d)   \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)

BÀi 2 : 

a)   \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)

\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)

b)   \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)

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

c)  \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)

\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)

\(=\left(b+c-a\right)\left(d-c^2\right)\)

BÀi 3 : 

a)  \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)

b)  \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)

c)   \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)

\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)

d)   \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\)  \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)

1 tháng 10 2020

a) \(5ax-15ay+20a\)

\(=5a\left(x-3y+4\right)\)

b) \(6xy-12x-8y\)

\(=6\left(xy-2x-3y\right)\)

c) \(3ab\left(x-y\right)+3a\left(y-x\right)\)

\(=3a\left(x-y\right)\left(b-1\right)\)

d) \(x^2-xy+2x-2y\)

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

1 tháng 10 2020

e) \(ax^2-5x^2-ax+5x+a-5\)

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

1 tháng 10 2020

a, \(5ax-15ay+20a=5a\left(x-5y+4\right)\)

b, sai 

c, \(3ab\left(x+y\right)+3a\left(y-x\right)=3ab\left(x+y\right)-3a\left(x+y\right)=\left(3ab-3a\right)\left(x+y\right)\)

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

Tượng tự ... 

1 tháng 10 2020

a) 5ax - 15ay + 20a = 5a( x - 3y + 4 )

b) 6xy - 12x - 8y = 2( xy - 6x - 4y )

c) 3ab( x - y ) + 3a( y - x ) = 3ab( x - y ) - 3a( x - y ) = ( x - y )( 3ab - 3a ) = 3a( x - y )( b - 1 )

d) x2 - xy + 2x - 2y = x( x - y ) + 2( x - y ) = ( x - y )( x + 2 )

e) ax2 - 5x2 - ax + 5x + a - 5 = x2( a - 5 ) - x( a - 5 ) + ( a - 5 ) = ( a - 5 )( x2 - x + 1 )

g) x2y - 4xy2 + 4y3 - 36yz2 = y( x2 - 4xy + 4y2 - 36z2 ) = y[ ( x2 - 4xy + 4y2 ) - 36z2 ] = y[ ( x - 2y )2 - ( 6z )2 ] = y( x - 2y - 6z )( x - 2y + 6z )

h) 4xy - x2 - 4y2 + m2 - 6m + 9

= ( m2 - 6x + 9 ) - ( x2 - 4xy + 4y2 )

= ( m - 3 )2 - ( x - 2y )2

= ( m - 3 - x + 2y )( m - 3 + x - 2y )

i) x2 + x - 12 = x3 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )

k) 5x2 + 14x - 3 = 5x2 - x + 15x - 3 = x( 5x - 1 ) + 3( 5x - 1 ) = ( 5x - 1 )( x + 3 )

m) x2 - 5xy + 4y2 = x2 - xy - 4xy + 4y2 = x( x - y ) - 4y( x - y ) = ( x - y )( x - 4y ) < đã sửa đề >

n) 3x2 - 5xy + 2y2 + 4x - 4y = ( 3x2 - 5xy + 2y2 ) + ( 4x - 4y ) = ( 3x2 - 3xy - 2xy + 2y2 ) + 4( x - y ) = [ 3x( x - y ) - 2y( x - y ) ] + 4( x - y ) = ( x - y )( 3x - 2y ) + 4( x - y ) = ( x - y )( 3x - 2y + 4 )

f) 2x3 + 4x2y + 2xy2 = 2x( x2 + 2xy + y2 ) = 2x( x + y )2

24 tháng 10 2020

a) ax + ay - bx - by = ( ax - bx ) + ( ay - by ) = x( a - b ) + y( a - b ) = ( a - b )( x + y ) < đã sửa >

b) 2x2 - 6xy + 5x - 15y = 2x( x - 3y ) + 5( x - 3y ) = ( x - 3y )( 2x + 5 )

c) ( a + b )2 - 4a2 = ( a + b )2 - ( 2a )2 = ( a + b - 2a )( a + b + 2a ) = ( b - a )( b + 3a )

d) 5a2xy - 10a3x - 15a2x2 = 5a2x( y - 2a - 3x )

e) 3( x - 1 ) + 5x( x - 1 ) = ( x - 1 )( 3 + 5x )

f) 9a2 - 4 = ( 3a )2 - 22 = ( 3a - 2 )( 3a + 2 )

g) 2x3 + 8x4 + 8x = 2x( x + 4x2 + 4 ) 

h) a2 - 4 + 4b - b2 = a2 - ( b2 - 4b + 4 ) = a2 - ( b - 2 )2 = ( a - b + 2 )( a + b - 2 )

i) a2 + 2ab + b2 - 16 = ( a2 + 2ab + b2 ) - 16 = ( a + b )2 - 42 = ( a + b - 4 )( a + b + 4 )

k) x2 + 5x + 4 = x2 + x + 4x + 4 = x( x + 1 ) + 4( x + 1 ) = ( x + 1 )( x + 4 )

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

m) x3 + 6x2 + 9x = x( x2 + 6x + 9 ) = x( x + 3 )2

27 tháng 10 2018

\(1.\)

\(a.\)

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

b.

\(3y^3+6xy^2+3x^2y\)

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

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

\(c.\)

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

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

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

\(2.\)

\(a.\)

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

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

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

\(b.\)

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

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

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

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

\(c.\)

\(x^2-6xy+9y^2-16\)

\(=\left(x^2-6xy+9y^2\right)-4^2\)

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

\(=\left(x-3-4\right)\left(x-3+4\right)\)

\(=\left(x-7\right)\left(x+1\right)\)

Tương tự câu \(d,e,g\)

\(3.\)

\(a.\)

\(x^3-2x=0\)

\(\Rightarrow x\left(x^2-2\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)

\(b.\)

\(x\left(x-4\right)+\left(x-4\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

\(c.\)

\(x\left(x-3\right)+4x-12=0\)

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

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

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

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

Tương tự \(d,e,g\)

27 tháng 10 2018

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

b)\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)

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

26 tháng 12 2018

1,4x2.(5x3+2x-1)

=4x2.5x3+4x2.2x-4x2.1

20x5+8x3-4x2

2,4x3y2:x2

=4xy2

3,(15x2y3-10x3y3+6xy):5xy

15x2y3:5xy-10x3y3:5xy+6xy:5xy

3xy2-2x2y2+\(\dfrac{6}{5}\)

26 tháng 12 2018

cảm ơn bạn nhé ^^