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19 tháng 7 2015

  3 - 6x + 3x^2  

 = 3 ( 1 - 2x + x^2 )

= 3( 1 - x )^2 

 

b, x^2 - 4xy + 4y^2 

= ( x)^2 + 2.x.2y + (2y)^2 

= ( x+ 2y)^2 

 

10 tháng 7 2018

a) xy – 3x + 2y – 6

= (xy - 3x) + (2y - 6)

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

= (y - 3)(x + 2)

b) x2y + 4xy + 4y – y3

= y(x2 + 4x + 4 - y2)

= y[(x2 + 4x + 4) - y2]

= y[(x + 2)2 - y2]

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

c) x2 + y2 + xz + yz + 2xy

= (x2 + 2xy + y2) + (xz + yz)

= (x + y)2 + z(x + y)

= (x + y)(x + y + z)

d) x3 + 3x2 – 3x – 1

= (x3 - 1) + (3x2 - 3x)

= (x - 1)(x2 + x + z) + 3x(x - 1)

= (x - 1)(x2 + 4x + 1)

10 tháng 7 2018

a ) 

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

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

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

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

b ) 

\(x^2y+4xy+4y-y^3\)

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

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

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

c ) 

\(x^2+y^2+xz+yz+2xy\)

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

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

7 tháng 10 2018

a) 2xy2 - 6x2y + 4xy

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

b) x2 - y2 - 5x + 5y

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

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

c) x2 - 4y2 - 1 + 4y

= x2 - (4y2 - 4y + 1)

= x2 - [ (2y)2 - 2.2.y.1 + 12 ]

= x2 - (2y-1)2

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

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 2016

a)x^2.16-4xy+4y^2

<=>16.x^2-2x2y+(2y)^2

<=>16(x-2y)^2

b)x^5-x^4+x^3-x^2

<=>(x^5-x^4)+(x^3-x^2)

<=>x^4(x-1)+x^2(x-1)

<=>(x-1)(x^4+x^2)

c)x^5+x^3-x^2-1

<=>(x^5+x^3)-(x^2+1)

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

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

d)x^4-3x^3-x+3

<=>(x^4-3x^3)-(x-3)

<=>x^3(x-3)-(x_3)

<=>(x-3)(x^3-1)

4 tháng 8 2016

\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)

 

12 tháng 12 2017

a. 2x-1-x2= -(x2-2x+1)=-(x-1)2

b. 8x3+y6=(2x)3+(y2)3

=(2x+y2)(4x2-2xy2+y4)

c. x2-16+4xy+4y2=(x2+4xy+4y2)-16

=(x+2y)2-16=(x+2y+4)(x+2y-4)

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)

24 tháng 9 2020

a) x3 + x2y - x2z - xyz

= ( x3 + x2y ) - ( x2z + xyz )

= x2( x + y ) + xz( x + y )

= ( x + y )( x2 + xz )

= x( x + y )( x + z )

b) x2 - y2 + 6x + 9

= ( x2 + 6x + 9 ) - y2

= ( x + 3 )2 - y2

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

c) x2 - 4xy - x + 2y + 4y2

= ( x2 - 4xy + 4y2 ) - ( x - 2y )

= ( x - 2y )2 - ( x - 2y )

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

d) 18x3 - 12x2 + 3x - 2

= ( 18x3 - 12x2 ) + ( 3x - 2 )

= 6x2( 3x - 2 ) + ( 3x - 2 )

= ( 3x - 2 )( 6x2 + 1 )

e) a2 + 2ab + b2 - c2 + 2cd - d2

= ( a2 + 2ab + b2 ) - ( c2 - 2cd + d2 ) 

= ( a + b )2 - ( c - d )2

= ( a + b - c + d )( a + b + c - d )

f) xz - yz - x2 + 2xy - y2

= z( x - y ) - ( x2 - 2xy + y2 )

= z( x - y ) - ( x - y )2

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

24 tháng 9 2020

a) x3 + x2y - x2z - xyz

= ( x3 + x2y ) - ( x2z + xyz )

= x2( x + y ) + xz( x + y )

= ( x + y )( x2 + xz )

= x( x + y )( x + z )

b) x2 - y2 + 6x + 9

= ( x2 + 6x + 9 ) - y2

= ( x + 3 )2 - y2

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

c) x2 - 4xy - x + 2y + 4y2

= ( x2 - 4xy + 4y2 ) - ( x - 2y )

= ( x - 2y )2 - ( x - 2y )

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

d) 18x3 - 12x2 + 3x - 2

= ( 18x3 - 12x2 ) + ( 3x - 2 )

= 6x2( 3x - 2 ) + ( 3x - 2 )

= ( 3x - 2 )( 6x2 + 1 )

e) a2 + 2ab + b2 - c2 + 2cd - d2

= ( a2 + 2ab + b2 ) - ( c2 - 2cd + d2 ) 

= ( a + b )2 - ( c - d )2

= ( a + b - c + d )( a + b + c - d )

f) xz - yz - x2 + 2xy - y2

= z( x - y ) - ( x2 - 2xy + y2 )

= z( x - y ) - ( x - y )2

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

1 tháng 11 2017

a,81-(x^2-4xy+4y^2)=81-(x-2y)^2=(9-(x-2y))(9+(x-2y))=(9-x+2y)(9+x-2y)

b,x^3+y^3+z^3-3xyz=(x^3+3(x^2)y+3x(y^2)+y^3)+z^3-3xyz-3xy(x+y)

=((x+y)^3+3((x+y)^2)z+3(x+y)z^2+z^3)-(3xyz-3xy(x+y))-3(x+y)z(x+y+z)

=(x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y+z)=(x+y+z)((x+y+z)^2-3(x+y)z-3xy)

=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xy-3yz-3xz)=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

3 tháng 9 2018

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

30 tháng 10 2019

Ta có:

a) 6x2y - 3y2 - 2x2 + y = (6x2y - 2x2) - (3y2 - y) = 2x2(3y - 1) - y(3y - 1) = (2x2 - y)(3y - 1)

b)  2x2 + x - 4xy - 2y + 2x + 1 = (x2 + x) - (4xy + 2y) + (x2 + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)2

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

c) 16x2y - 4xy2 - 4x3 + x2y = 4xy(4x - y) - x2(4x - y) = (4xy - x2)(4x - y)

d) 4x2 - 20x + 25 - 36y2 = (2x  - 5)2 - (6y)2 = (2x - 5 - 6y)(2x  - 5 + 6y)

e) x2 - 4y2 + 6x - 4y + 8 = (x2 + 6x + 9) - (4y2 + 4y + 1) = (x + 3)2 - (2y + 1)2 = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)

30 tháng 10 2019

g) Ta có : x10 + x5 + 1

= (x10 - x) + (x5 - x2) + (x2 + x + 1)

= x(x9 - 1) + x2(x3 - 1) + (x2 + x + 1)

= x(x3 - 1)(x6 + x3 + 1) + x2(x3 - 1) + (x2 + x + 1)

= (x7 + x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)(x2 + x + 1) + (x2 + x + 1)

= (x2 + x + 1)(x8 - x7 + x 5 - x4 + x2 - x + x4 + x3 + x2 + 1)

= (x2 + x + 1)(x8 - x7 + x5 + x3 - x + 1)

h) TT trên (dài dòng ktl)