\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)

Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)

\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)

\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)

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

giúp mềnh với mn ơi ;(((

a) \(x^2+4y^2+4xy\)

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

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

b) \(1-10x+25x^2\)

\(=\left(5x\right)^2-10x+1\)

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

c) \(x^2+4x^4-4xy^2\)

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

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

1: =(16x^2-8x+1)-y^2

=(4x-1)^2-y^2

=(4x-1-y)(4x-1+y)

2: =(x^2-2xy+y^2)-z^2

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

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

3: =(x^2+4xy+4y^2)-16

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

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

4: =(x^2-4xy+4y^2)-16

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

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

Đặt \(P=x^2+4y^2-4xy+2x-4y+9\)

\(P=\left(x-2y\right)^2+2\left(x-2y\right)+1+8\)

\(P=\left(x-2y+1\right)^2+8\ge8\)

\(P_{min}=8\) khi \(x-2y+1=0\)

\(D=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4y^2+4xy\right)-\left(2x+4y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay \(x+2y=5;\)có :

\(D=5^2-2.5+10\)

\(=25-10+10\)

\(=25\)

Vậy...