Giúpoppppppopp Bài 2:,$1)~(2x+1)^2+(2x-1)^2$,$2)~-(x+1)^2-(x-1)^2$,$3)~(x+2y)^2-(x-2y)^2$,$4)~(3x+y)^2+(x-y)^2$,$5)~-(x+5)^2-(x-8)^2$,$6)~(3x-2)^2-(3x-1)^2$,$7)~(x-4y)^2+(x+4y)^2$,$8)~-(-2x+3)^2-(5x-3)^2$,$9)~(-2x+3)^2-(5x-3)^2$

Question

Accepted Answer

Bài 2:
1) $(2x+1)^2+(2x-1)^2$
$$
= (2x)^2 + 2 \cdot 2x \cdot 1 + 1^2 + (2x)^2 - 2 \cdot 2x \cdot 1 + 1^2 
= 4x^2 + 4x + 1 + 4x^2 - 4x + 1 
= 8x^2 + 2
$$

2) $-(x+1)^2-(x-1)^2$
$$
= -(x^2 + 2x + 1) - (x^2 - 2x + 1) 
= -x^2 - 2x - 1 - x^2 + 2x - 1 
= -2x^2 - 2
$$

3) $(x+2y)^2-(x-2y)^2$
$$
= (x^2 + 4xy + 4y^2) - (x^2 - 4xy + 4y^2) 
= x^2 + 4xy + 4y^2 - x^2 + 4xy - 4y^2 
= 8xy
$$

4) $(3x+y)^2+(x-y)^2$
$$
= (9x^2 + 6xy + y^2) + (x^2 - 2xy + y^2) 
= 9x^2 + 6xy + y^2 + x^2 - 2xy + y^2 
= 10x^2 + 4xy + 2y^2
$$

5) $-(x+5)^2-(x-8)^2$
$$
= -(x^2 + 10x + 25) - (x^2 - 16x + 64) 
= -x^2 - 10x - 25 - x^2 + 16x - 64 
= -2x^2 + 6x - 89
$$

6) $(3x-2)^2-(3x-1)^2$
$$
= (9x^2 - 12x + 4) - (9x^2 - 6x + 1) 
= 9x^2 - 12x + 4 - 9x^2 + 6x - 1 
= -6x + 3
$$

7) $(x-4y)^2+(x+4y)^2$
$$
= (x^2 - 8xy + 16y^2) + (x^2 + 8xy + 16y^2) 
= x^2 - 8xy + 16y^2 + x^2 + 8xy + 16y^2 
= 2x^2 + 32y^2
$$

8) $-(-2x+3)^2-(5x-3)^2$
$$
= -((4x^2 - 12x + 9)) - (25x^2 - 30x + 9) 
= -4x^2 + 12x - 9 - 25x^2 + 30x - 9 
= -29x^2 + 42x - 18
$$

9) $(-2x+3)^2-(5x-3)^2$
$$
= (4x^2 - 12x + 9) - (25x^2 - 30x + 9) 
= 4x^2 - 12x + 9 - 25x^2 + 30x - 9 
= -21x^2 + 18x
$$