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

\(\left(xy-1\right)^2+\left(x+y\right)^2\)

\(=x^2y^2-2xy+1+x^2+2xy+y^2\)

\(=x^2y^2+x^2+y^2+1\)

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

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

=a³ + b³ ( hdt số 6)

\(\left(\text{a}^3+b^3\right)\)

a)Ta có \(3x\left(x+1\right)-5y\left(x+1\right)=\left(x+1\right)\left(3x-5y\right)\)

b)Ta có \(3x^3\left(2y-3z\right)-15x\left(2y-3z\right)=\left(2y-3z\right)\left(3x^3-15x\right)=\left(2y-3z\right)3x\left(x^2-5\right)\)

k)Ta có \(3x\left(z+2\right)+5\left(-z-2\right)=\left(z+2\right)\left(3x-5\right)\)

c)Ta có \(4y\left(x-1\right)-\left(1-x\right)=\left(x-1\right)\left(4y+1\right)\)

\(=2y^3+2y^2+2y-2x^3-2y^2-2y-20=-20\)

`H = 2y^3 + 2y^2 + 2y - 2y^3 - 2y^2 - 2y - 20`

`= (2y^3 - 2y^3) + (2y^2 -2 y^2) + (2y-2y) - 20`

`= -20`.

`=>` Bth ko pth vào biến,

`H = 2y(y^2 + y + 1)- 2y^2(y+1)-2.(y+10)`

`H = 2y^3+2y^2 +2y - 2y^3 - 2y^2 - 2y - 20`

`H = ( 2y^3 - 2y^3) + ( 2y^2 -2y^2) + (2y-2y) - 20`

`H = -20`

`=>` Biểu thức trên ko phụ thuộc vào biến `x (đfcm)`

\(\text{Δ}=16^2-4\cdot\left(-3\right)\cdot\left(-297\right)=-3308< 0\)

Do đó: PTVN

`-> -3(x^2 + 16/3 + 99) = 0`.

`-> -3(x^2 + 2 . 8/3 + 64/9 + 827/9) =0`

`-> -3(x-8/3)^2 - 827/3 = 0`.

Vì `-3(x-8/3)^2 <=0 => -3(x-8/3)^2 - 827/3 <= -827/3 < 0`.

`=> x = cancel O`.

Đk: `x ne 0, -3`.

Với `x` thỏa mãn thì ptr tương đương với;
`((33x - 99) - 62x)/(x(x+3)) = 3/2`

`=> (-19x - 99)/(x(x+3)) = 3/2`

`=> -38x - 198 = 3x^2 + 9x`.

`<=> 3x^2 + 47x + 198 =0`

`=> 3(x^2 + 2 .47/6x + 2009/54 + 8693/54) = 0`.

`<=> 3(x + 47/6)^2 + 8693/18 = 0` 

`=> x = cancel O`.