a: =(x^2y-x^3)-(9y-9x)

=x^2(y-x)-9(y-x)

=(y-x)(x^2-9)

=(y-x)(x-3)(x+3)

b: \(=\left(x^2-2xy+y^2\right)-4\)

=(x-y)^2-4

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

c: \(=\left(x^2+4x+4\right)-y^2\)

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

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

d: =(x^2-y^2)-(2x+2y)

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

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

\(a,x^2y-x^3-9y+9x\)

\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(y-x\right)\left(x^2-9\right)\)

\(=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)

\(b,x^2-2xy+y^2-4\)

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

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

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

\(c,x^2+4x-y^2+4\)

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

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

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

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

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

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

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

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

#Urushi

đề sai

C1: \(\left(x+y\right)\left(x-y\right)=x\left(x-y\right)+y\left(x-y\right)=x^2-xy+xy-y^2=x^2-y^2\)

C2: x²-y²=(x-y)(x+y)

  <=> x²-y²-(x-y)(x+y)=0

   <=> x²-y²-[x(x+y)-y(x+y)] = 0

   <=> x²-y²-(x²+xy-xy-y²) = 0

    <=> x²-y²-(x²-y²) = 0

    <=> x²-y²-x²+y² = 0

    <=> 0 =0 (đúng)

Vậy .....

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

Co ( x + y )( x - y ) = x^2 - xy + xy - y^2 = x^2 - y^2

Ma x^2 - y^2 = x^2 - y^2

=> x^2 - y^2 = ( x + y )( x - y ) 

Ói , hoa mắt chóng mặt nhức đầu ,

Áp dụng BĐT Cauchy-Schwaz: 

\(\left(\frac{x^3}{y^2}+\frac{9y^2}{x+2y}\right)\left[xy^2+y^2\left(x+2y\right)\right]\ge\left(x^2+3y^2\right)^2\)

\(\Leftrightarrow\frac{x^3}{y^2}+\frac{9y^2}{x+2y}\ge\frac{\left(x^2+3y^2\right)^2}{2xy^2+2y^3}\)

\(\Leftrightarrow\frac{x^3}{y^2}+\frac{9y^2}{x+2y}\ge\frac{\left(x^2+3y^2\right)^2}{2y^2\left(x+y\right)}\)        \(\left(1\right)\)

 Áp dụng BĐT AM-GM:

\(x^2+y^2\ge2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow\left(x^2+y^2\right)^2\ge\left(x+y\right)^2\)

\(\Rightarrow x^2+y^2\ge x+y\)           

Do đó: Áp dụng BĐT AM-GM ngược dấu: 

   \(2y^2\left(x+y\right)\le2y^2\left(x^2+y^2\right)\le\frac{\left(x^2+y^2+2y^2\right)^2}{4}\)

\(\Leftrightarrow2y^2\left(x+y\right)\le\frac{\left(x^2+3y^2\right)^2}{4}\)               \(\left(2\right)\)

Từ (1) và (2) suy ra \(\frac{x^3}{y^2}+\frac{9y^2}{x+2y}\ge4\)   (đpcm)

Dấu "=" xảy ra khi x=y=1

Vậy \(\frac{x^3}{y^2}+\frac{9y^2}{x+2y}\ge4\)

a.

\(x^2+4y^2+4xy=0\)

\(\Leftrightarrow\left(x+2y\right)^2=0\)

\(\Leftrightarrow x+2y=0\)

\(\Leftrightarrow x=-2y\)

Vậy pt đã cho có vô số nghiệm dạng \(\left(x;y\right)=\left(-2k;k\right)\) với k là số thực bất kì (nếu đề đúng)

b.

\(2y^4-9y^3+2y^2-9y=0\)

\(\Leftrightarrow2y^2\left(y^2+1\right)-9y\left(y^2+1\right)=0\)

\(\Leftrightarrow\left(2y^2-9y\right)\left(y^2+1\right)=0\)

\(\Leftrightarrow y\left(2y-9\right)\left(y^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\\2y-9=0\\y^2+1=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{9}{2}\end{matrix}\right.\)

c. Em kiểm tra lại đề chỗ \(3xy^2\), đề đúng như vậy thì pt này ko giải được

\(a,=2\left(xy^2-2\right)\\ b,=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\\ c,=y\left(x^2-6x-9\right)\\ d,=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)