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\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y-x+y+2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2}{x-y}\)

2 tháng 1 2022

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

18 tháng 7 2021

a, \(\dfrac{x^2}{4}-xy+y^2=\left(\dfrac{x}{2}\right)^2-xy+y^2=\left(\dfrac{x}{2}\right)^2-2.\dfrac{x}{2}.y+y^2\)

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

b, \(x^2+x+\dfrac{1}{4}=x^2+\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

c, \(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}x+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

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

 

18 tháng 7 2021

`x^2/4-2*x/2*y+y^2`

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

`x^2+x+1/4`

`=x^2+2*x*1/2+(1/2)^2`

`=(x+1/2)^2`

`x^2+2sqrt3x+3`

`=x+2xsqrt3+sqrt3^2`

`=(x+sqrt3)^2`

`4x^2-1`

`=(2x)^2-1`

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

20 tháng 4 2017

Bài giải:

a) 3x - 6y = 3 . x - 3 . 2y = 3(x - 2y)

b) 2525x2 + 5x3 + x2y = x2 (2525 + 5x + y)

c) 14x2y – 21xy2 + 28x2y2 = 7xy . 2x - 7xy . 3y + 7xy . 4xy = 7xy(2x - 3y + 4xy)

d) 2525x(y - 1) - 2525y(y - 1) = 2525(y - 1)(x - y)

e) 10x(x - y) - 8y(y - x) =10x(x - y) - 8y[-(x - y)]

= 10x(x - y) + 8y(x - y)

= 2(x - y)(5x + 4y)

4 tháng 9 2017

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

b,\(x^2(\dfrac{2}{5}+5x+y)\)

c,\(7xy\left(2x-3y+4xy\right)\)

d,\(\dfrac{2}{5}x\left(y-1\right)-\dfrac{2}{5}y\left(y-1\right)\)

=\(\dfrac{2}{5}\left(y-1\right)\left(x-y\right)\)

e,\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)\)

\(2\left(x-y\right)\left(5x+4y\right)\)

24 tháng 9 2019

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

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

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

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

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

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

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

24 tháng 9 2019

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

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

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

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

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

.................

2 tháng 1 2022

đáp án: a là đúng

`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`

`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`

`c, xy^2 + x^2y + 1/4y^3`

`= y(xy + x^2 + 1/4y^2)`

`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`

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

4 tháng 7 2017

1, \(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}+\dfrac{y}{y-x}\)

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

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

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

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

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

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

2, \(x^2-y^2-2y-1\)

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

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

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