\(B=4x^2+6x^2y^2+2y^4+20y^2\)

\(=4x^2+4x^2y^2+2x^2y^2+2y^4+20y^2\)

\(=4x^2.\left(x^2+y^2\right)+2y^2.\left(x^2+y^2\right)+20y^2\)

\(=\left(x^2+y^2\right).\left(4x^2+2y^2\right)+20y^2\)

Bí >>>

a, \(M=\frac{xy^2+y^2\left(y^2-x\right)+1}{x^2y^4+2y^4+x^2+2}=\frac{y^2\left(x+y^2-x\right)+1}{y^4\left(x^2+2\right)+\left(x^2+2\right)}=\frac{y^4+1}{\left(y^4+1\right)\left(x^2+2\right)}=\frac{1}{x^2+2}\)

Thay x=-3 vào M

=>\(M=\frac{1}{\left(-3\right)^2+2}=\frac{1}{11}\)

b, Vì \(x^2\ge0\Rightarrow x^2+2\ge2\Rightarrow M=\frac{1}{x^2+2}>0\)

Lời giải:

$M=4x^2(x^2+y^2)+2y^2(x^2+y^2)+20y^2$

$=4x^2.10+2y^2.10+20y^2$

$=40x^2+20y^2+20y^2=40x^2+40y^2=40(x^2+y^2)=40.10=400$

\(N=3x^4+3x^2y^2+x^2y^2+y^4+2y^2\)

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

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

a ) A = M + N = ( 2x²y - xy² + 3x - 2y ) + ( 2xy² - 2x²y - 5x + 2y )

                      =  2x²y - xy² + 3x - 2y + 2xy² - 2x²y - 5x + 2y 

                      =  ( 2x²y - 2x²y ) + ( -xy² + 2xy² ) + ( 3x - 5x ) + ( - 2y + 2y )

                      =   0 + ( -1 +2 ) xy² + ( 3 - 5 )x + 0

                      =  xy² - 2x

     Vậy A = M + N = xy² - 2x

    B = N - M =  2xy² - 2x²y - 5x + 2y - ( 2x²y - xy² + 3x - 2y )

                    =    2xy² - 2x²y - 5x + 2y - 2x²y + xy² - 3x + 2y 

                    =  ( 2xy² + xy² ) + ( -2x²y - 2x²y ) + ( - 5x - 3x ) + (  2y + 2y )

                    =  ( 2 + 1 )xy² + ( -2 - 2 )x²y  + ( - 5 - 3 )x  + (  2 + 2 )y 

                    =  3xy² - 4x²y  - 8x  + 4y 

 Vậy B = 3xy² - 4x²y  - 8x  + 4y 

a) ĐKXĐ: \(x\ne2y,x\ne-y;x\ne-1\) 

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

\(B=\left[\dfrac{y-x}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{4x^4+4x^2y+y^2-4}{x\left(x+y\right)+\left(x+y\right)}\)

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

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

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

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

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