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21 tháng 12 2016

a) 10x(x-y)-6y(y-x)=10x(x-y)+6y(x-y)=(10x+6y)(x-y)

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

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

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

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

d)\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)\(=\left(x+3\right)\left(x+1\right)\)

e)\(x^2-4x-5=x^2-5x+x-5=x\left(x-5\right)+\left(x-5\right)\)\(=\left(x+1\right)\left(x-5\right)\)

21 tháng 12 2016

dễ quá

8 tháng 9 2018

câu 1 

a, 5x - x 2 + 2xy - 5y 

= 5x - x 2 + xy + xy - 5y 

= ( 5x - 5y ) - ( x2 - xy ) + xy 

= 5 ( x-y ) - x(x-y ) + xy 

= (5-x) ( x-y) + xy 

mik làm dc mỗi câu a ! 

16 tháng 12 2018

Câu 1:

a/ (-5x3)(2x2+3x-5)

=-10x5-15x4+25x3

b/(2x-1)x

=2x2-x

c/(x-y)(3x2+4xy)

=3x3+4x2y-3x2y-4xy2

=3x3 +x2y-4xy2

Câu 2:

a/ x3-2x2+x

=x(x2-2x+1)

=x(x-1)2

b/x2-x-12

=x2 +3x-4x-12

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

=x(x+3)-4(x+3)

=(x+3)(x-4)

c/ 2x-6

=2(x-3)

e/ x2+4x+4-y2

=(x2+4x+4)-y2

=(x+2)2-y2

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

d/ x2-2xy+y2-16

=(x2-2xy+y2)-16

=(x-y)2-16

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

Câu 3:

a: \(=\dfrac{5xy-4+3xy+4}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)

b: \(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)

c: \(=\dfrac{3x+1-2x+3}{x+y}=\dfrac{x+4}{x+y}\)

d: \(=\dfrac{4x+7+5x+7}{9}=\dfrac{9x+14}{9}\)

e: \(=\dfrac{5\left(x+2\right)}{2\left(2x-1\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-5\left(x-2\right)}{2x-1}\)

27 tháng 10 2017

a) \(=2xy^2\left(x^2+8x+15\right)\)

\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)

\(=2xy^2\left[\left(x+4\right)^2-1\right]\)

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

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

mấy câu sau tự làm nha :*

29 tháng 10 2017

b,=(x^2-10x+25)-4

  =(x-5)^2-2^2

  =(x-5-2)(x-5+2)

  =(x-7)(x-3)

4 tháng 12 2017

a, = (x^2+10x+25)-y62 = (x+5)^2-y^2 = (x+5-y).(x+5+y)

b, = xy.(x-y)

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

k mk nha

28 tháng 1 2020

Thank you.

27 tháng 9 2017

\(a,2x^2-2xt-5x+5y\)

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

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

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

\(b,8x^2+4xy-2ax-ay\)

\(=\left(8x^2-2ax\right)+\left(4xy-ay\right)\)

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

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

\(c,x^3-4x^2+4x\)

\(=x^3-2x^2-2x^2+4x\)

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

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

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

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

\(d,2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

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

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

\(e,x^2-y^2-2yz-z^2\)

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

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

1 tháng 11 2016

a) (x^2+x)^2-14(x^2+x)+24

=(x^2+x)^2-2(x^2+x)-12(x^2+x)24

=(x^2+x)(x^2+x-2)-12(x^2+x-2)

=(x^2+x-12)(x^2+x-2)

1 tháng 11 2016

b)x^2-y^2+4-4x

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

=(x-2)^2-y^2

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

3 tháng 9 2018

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

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

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

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

1 tháng 10 2020

\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\  = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\  = {a^2}{b^2} + 1 + {a^2} + {b^2}\\  = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\  = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\  = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\  = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\  = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\  = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\  = {x^3} + 2{x^2} + x + x + 1\\  = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\  = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\  = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\  = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\  = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\  = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\  = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\  = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\  = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\  = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\  = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\  = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\  = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array}