K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 7 2018

a)  \(C=ax-ay-bx+by+\left(y-x\right)^2\)

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

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

b)  \(M=\left(5x-10\right)\left(x^2-1\right)\left(3x-6\right)\left(x^2-2x+1\right)\)

\(=5\left(x-2\right)\left(x-1\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)^2\)

\(=\left(x-2\right)\left(x-1\right)\left[5\left(x+1\right)-3\left(x-1\right)\right]\)

\(=\left(x-2\right)\left(x-1\right)\left(5x+5-3x+3\right)\)

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

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

20 tháng 7 2018

còn mà bạn

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

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

4 tháng 8 2017

Mình sửa: Bài 1
2)x2+3x-15

20 tháng 5 2018

a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2

b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)

                         = -(52 – 2 . 5 . x – x2) = -(5 – x)2

c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]

                    = (2x - 1/2)(4x2 + x + 1/4) 

d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)

8 tháng 9 2020

A = xy + y - 2x - 2

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

= ( x + 1 )( y - 2 )

B = x2 - 3x + xy - 3y

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

= ( x - 3 )( x + y )

C = 3x2 - 3xy - 5x + 5y

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

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

D = xy + 1 + x + y

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

= ( x + 1 )( y + 1 )

E = ax - bx + ab - x2

= ( ax - x2 ) + ( ab - bx )

= x( a - x ) + b( a - x )

= ( a - x )( x + b )

F = x2 + ab + ax + bx

= ( ax + x2 ) + ( ab + bx )

= x( a + x ) + b( a + x )

= ( a + x )( x + b )

G = a3 - a2x - ay + xy

= a2( a - x ) - y( a - x )

= ( a - x )( a2 - y )

Bonus : = ( a - x )[ a2 - ( √y )2 ]

             = ( a - x )( a - √y )( a + √y )

H = 2xy + 3z + 6y + xz

= ( 6y + 2xy ) + ( 3z + xz )

= 2y( 3 + x ) + z( 3 + x )

= ( 3 + x )( 2y + z )

8 tháng 9 2020

A = xy + y - 2x - 2 = y(x + 1) - 2(x + 1) = (y - 2)(x + !1

B = x2 - 3x + xy - 3y = x(x - 3) + y(x - 3) = (x + y)(x - 3)

C = 3x2 - 3xy - 5x + 5y = 3x(x - y) - 5(x - y) = (3x - 5)(x - y)

D = xy + 1 + x + y = xy + x + y + 1 = x(y + 1) + (y + 1) = (x + 1)(y + 1)

E = ax - bx + ab - x2 = ax - x2 + ab - bx = a(a - x) - b(a - x) = (a - b)(a - x)

F = x2 + ab + ax + bx = ab + ax + bx + x2 = a(b + x) + x(b + x) = (a + x)(b + x)

G = a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)

H = 2xy + 3z + 6y + xz = 2xy + 6y + 3z + xz = 2y(x + 3) + z(x + 3) = (2y + z)(x + 3)

15 tháng 10 2017

a) ko bt làm

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

9 tháng 12 2018

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

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

\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)

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

\(=\left(x-3\right)\left(2x^2-11x\right)\)

\(=x\left(x-3\right)\left(2x-11\right)\)

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

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

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

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

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

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

9 tháng 12 2018

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

\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)

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

 \(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)

14 tháng 7 2019

a) \(4x^2-4x+1=\left(2x-1\right)^2\)

14 tháng 7 2019

\(3x\left(x-5\right)-x\left(4+3x\right)=43\)

\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)

\(\Leftrightarrow-19x=43\)

\(\Leftrightarrow x=\frac{-43}{19}\)

Bài làm

a) 4x2 - 6x 

= 2x( 2x - 3 )

b) 9x4y3 + 3x2y

= 3x2y3( 3x2 + y )

c) x- 2x+ 5x

= x( x2 - 2x + 5 )

d) 3x( x - 1 ) + 5( x - 1 )

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

e) 2x2( x + 1 ) + 4( x + 1 )

= ( x + 1 )( 2x2 + 4 )

= ( x + 1 )2( x2 + 2 )

= 2( x + 1 )( x2 + 2 )

f) -3x - 6xy + 9xz

= -( 3x + 6xy - 9xz )

= -3x( 1 + 2y - 3z )

# Học tốt #