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2 tháng 8 2018

A=26x2+y(2x+y)-10x(x+y)

A=26x2+2xy+y2-10x2-10xy

A=16x2-8xy+y2 =(4x)2-2.4x.y+y2 =(4x-y)2

Thay x=0,25y,ta có: A=(4.0,25y - y)2=(y-y)2=0

B=x3+6x2y+12xy2+8y3

B=x3+3x22y+3x(2y)2+(2y)3 =(x+2y)3

Có x+2y=-5 ⇒ x=-5-2y

Thay x=-5-2y vào, ta có B=(-5-2y+2y)3=(-5)3=-125

20 tháng 8 2016

b)B=27y^3-27y^2x+9yx^2-x^3 
= 27 . (1/3x)^3 - 27.(1/3x)².x + 9.1/3.x.x^2 - x^3 
= x^3 - 3x^3 + 3x^3 - x^3 
= 0

d) D=50y^2+x(x-2y)+14y(x-y) 

=50y^2 +x^2 -2xy +14xy -14y^2 

=36y^2 +x^2 +12xy 

=(6y + x)^2 

=81 

1: \(=\dfrac{x^2-1}{x\left(x^2-1\right)}=\dfrac{1}{x}\)

2: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{y\left(x-2\right)}=\dfrac{x+2}{y}\)

3: \(=\dfrac{2x^2+2xy-xy-y^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)\left(2x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x-y}{x-y}\)

4: \(=\dfrac{x\left(x^2-1\right)}{x\left(x^2-x-2\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{x-1}{x-2}\)

 

a: \(=\dfrac{5x^2y^4}{-10x^2y}=-\dfrac{1}{2}y^3=-\dfrac{1}{2}\cdot8=-4\)

b: \(=\dfrac{15x^4y^2}{5x^3y}+\dfrac{20x^3y^2}{5x^2y}=3xy+4xy=7xy\)

\(=7\cdot\dfrac{1}{7}\cdot2009=2009\)

 

12 tháng 9 2018

1,\(\dfrac{x^2-6x+9}{x^2-8x+15}=\dfrac{\left(x-3\right)^2}{\left(x-3\right).\left(x-5\right)}=\dfrac{x-3}{x-5}\)

2,\(\dfrac{x^2+5x}{2x+10}=\dfrac{x.\left(x+5\right)}{2.\left(x+5\right)}=\dfrac{x}{2}\)

3,\(\dfrac{25-10x+x^2}{xy-5y}=\dfrac{\left(x-5\right)^2}{y.\left(x-5\right)}=\dfrac{x-5}{y}\)

4,\(\dfrac{x^2+3x-y^2-3y}{x^2-y^2}\\ \\ =\dfrac{\left(x+y\right).\left(x-y\right)+3.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}\\ \\ =\dfrac{\left(x-y\right).\left(x+y+3\right)}{\left(x-y\right).\left(x+y\right)}\\ \\ =\dfrac{x+y+3}{x+y}\)5,\(\dfrac{x^3+2x^2-x-2}{x^3-3x+2}=\dfrac{x^2.\left(x+2\right)-\left(x+2\right)}{x.\left(x^2-1\right)-2.\left(x-1\right)}\\ \\ \dfrac{\left(x+2\right).\left(x^2-1\right)}{x.\left(x+1\right).\left(x-1\right)-2.\left(x-1\right)}\\ =\dfrac{\left(x+2\right).\left(x+1\right).\left(x-1\right)}{\left(x-1\right).\left[\left(x+1\right).x-2\right]}=\dfrac{\left(x+2\right).\left(x+1\right)}{\left(x+1\right).x-2}\)

2 tháng 8 2018

1.x2-y2+2x+1=(x2+2x+1)-y2=(x+1)2-y2=(x+1-y)(x+1+y)

2.(x2+9)2-36x2=(x2+9)2-(6x)2=(x2+9-6x)(x2+9+6x)=(x-3)2(x+3)2

3.\(8x^3+\dfrac{1}{27}=\left(2x\right)^3+\left(\dfrac{1}{3}\right)^3\\ =\left(2x+\dfrac{1}{3}\right)\text{[}\left(2x\right)^2-2x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\text{]}\\ =\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)4.x3-8y3=x3-(2y)3=(x-2y)(x2+2xy+4y2)

3 tháng 3 2022

a) (x-y)2

b) (x-y)3

c) x+5y

d) x.(4+y)

e) (2k+1)2+(2k+3)2

sorry nha mình chỉ bt đến đây thôi

7 tháng 3 2022

a) \(\left(x-y\right)^2\)

b) \(\left(x-y\right)^3\)

c)  \(x+5y\)

d) \(x.\left(4+y\right)\)

e) \(\left(2k+1\right)^2+\left(2k+3\right)^2\)

f)    \(a+\frac{1}{a}\)\(\left(a\inℚ;a\ne0\right)\)

g)    \(\left(2k\right)^2+\left(2k+2\right)^2\)

24 tháng 6 2018

Giải:

1) \(a^3-3a^2+3a-1\)

\(=a^3-3a^2.1+3a.1^2-1^3\)

\(=\left(a-1\right)^3\)

Vậy ...

2) \(x^3+6x^2+12x+8\)

\(=x^3+3.x^2.2+3.x.2^2+2^3\)

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

Vậy ...

3) \(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)

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

Vậy ...

4) \(x^3-6x^2y+12xy^2-8y^3\)

\(=x^3-3.x^2.2y+3.x.\left(2y\right)^2-\left(2y\right)^3\)

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

Vậy ...

22 tháng 4 2018

a) A = x^3 + 6x^2y + 12xy^2 + 8y^3

=> A = ( x + 2y )^3

Thay x + 2y = -5 vào A

=> A = ( -5 )^3 = -125

Vậy khi x + 2y = -5 thì A = -125

b) B = 8x^3 - 12x^2y + 6xy^2 - y^3

=> B = ( 2x - y )^3

Thay 2x - y = 1/5 vào A

=> A = ( 1/5 )^3 = 1/125

Vậy khi 2x - y = 1/5 thì B = 1/125

c) C = x^3 + 3x^2 + 3x + 1

=> C = ( x + 1 )^3

Thay x = 99 vào C

=> C = ( 99 + 1 )^3 = 100^3 = 1000000

Vậy khi x = 99 thì C = 1000000