Cho: ${\left(x^2+y^2-17\right)}^2-4{\left(xy-4\right)}^2=\left(x+y+5\right)\left(x-y+3\right)\left(x+y+m\right)\left(x-y+n\right)$

Khi đó giá trị của m.n là

A. $x^2 – 6x + 9 = {(x – 3)}^2$

B. $4x^2 – 4xy + y^2 = {(2x – y)}^2$

C. $x^2+x+\frac{1}{4}={\left(x+\frac{1}{2}\right)}^2$

D. $-x^2 – 2xy – y^2 = -{(x – y)}^2$

 A. A = 1

B. A = 0

C. A = 2

D. Chưa đủ dữ kiện để tính

A. ${(5x – 4)}^2 – 49x^2 = -8(3x + 1)(x + 2)$

B. ${(5x – 4)}^2 – 49x^2 = (3x – 1)(x + 2)$

C. ${(5x – 4)}^2 – 49x^2 = -8(3x - 1)(x - 2)$

D. ${(5x – 4)}^2 – 49x^2 = -8(3x - 1)(x + 2)$

A. 9$x^2$ + 0,03x + 0,1

B. 9$x^2$ + 0,6x + 0,01

C. 9$x^2$ + 0,3x + 0,01

D. 9$x^2$ – 0,3x + 0,01

A. $4x^2 + 4x + 1 = {(2x + 1)}^2$

B. $9x^2 – 24xy + 16y^2 = {(3x – 4y)}^2$

C. $\frac{x^2}{4}+2xy+4y^2={\left(\frac{x}{2}+2y\right)}^2$

D. $\frac{x^2}{4}+2xy+4y^2={\left(\frac{x}{4}+2y\right)}^2$

A. ${(xy + 2)}^3$

B. ${(xy + 8)}^3$

C. $x^3{y}^3 + 8$

D. ${\left(x^3{y}^3+2\right)}^3$