1.A

2.C

3.B

4.C

a

c

b

c

ai júp với, tớ T I C K cho mỗi ngày, gấp lắm

ko ai jup sao, hic hic

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

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

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

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

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

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

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

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

2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3

3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3

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

1) x² + 4xy + 4y² = ( x + 2y )²

2) - x³ + 9x² - 27x + 27 = ( 3 - x )²

3) 8x⁶ + 36x⁴y + 54x²y² + 27y³ = ( 2x² + 3y )³

4) x³ - 6x²y + 12xy² - 8y³ = ( x - 2y )³

5) x² + 4y² +1 - 4xy - 2x + 4y = ( x² - 2y - 1 )²

6) x² + y² + 4 + 2xy + 4x + 4y = ( x + y + 2 )²

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

Câu 9: D

Câu 10: A

Ta có: \(x^2+4y^2+x=4xy+2y+2\)

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

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

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

Tìm các TH

Mặt khác : \(4x^2+4xy+y^2=2x+y+56\) 

                \(\Rightarrow\left(2x+y\right)^2-\left(2x+y\right)=56\)

               \(\Rightarrow\left(2x+y\right)\left(2x+y-1\right)=56\)

Tìm các TH