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30 tháng 3 2021

Ta có : 

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

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

Từ (1) và (2)

=> A > B

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

25 tháng 2 2021

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

`=(y(y-2x))/3`

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

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

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

`=(-x(x-3)+x-3)/((x-1)(x+1))`

`=((x-3)(1-x))/((x-1)(x+1))`

`=(3-x)/(1+x)`

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

\(=\dfrac{y\left(y-2x\right)}{3}\)

 

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

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

30 tháng 7 2023

a. (2x+3y)2= (2x)2+2.2x.3y+(3y)2

=4x2+12xy+9y2

b. 2(\(\dfrac{1}{2}\)x2+y)(x2-2y)

=(x2+2y)(x2-2y)

=x4-4y2

c, (x+y+z)2= [(x+y)+z]2

=(x+y)2+2(x+y)z+z2

=x2+2xy+y2+2xz+2yz+z2

=x2+y2+z2+2xy+2yz+2xz

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

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

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

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

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

14 tháng 12 2018

\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)

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

\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)

\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)

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

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