2/calculator

3/ book

4/ bag

5/ rubber

6/pencil case

c) Thay x= -1 vào đa thức P(x)

P(x)= -3x⁴ +8x²+5x = -3 x (-1 )⁴+ 8 x (-1 )²+ 5 x -1= -3 x 1 + 8 x 1 + 5 x -1 = -3 +8 -5 = 0

 A(x)= -x³-2x²+5x+7

 B(x)= -3x⁴ +x³+10x² -7

b) P(x) = A(x)+B(x)= -x³-2x²+5x+7-3x⁴ +x³+10x² -7=  -3x⁴ +8x²+5x

    Q(x)=  -x³-2x²+5x+7- (-3x⁴ +x³+10x² -7)=  -x³-2x²+5x+7 + 3x⁴-x³ - 10x² + 7= -2x³-12x²+5x+ 14

4) 

a/ A(x)= -4⁵-x³+4x²+ 5x+9+4x⁵-6x²-2

    A(x)= -x³-2x²+5x+7

b/ B(x)= -3x⁴-2x³ +10x² -8x+5x³-7-2x³+8x

    B(x)= -3x⁴ +x³+10x² -7

cho mình hỏi là bạn có ghi sai đề hok ạ? tại vì có AD rồi, nhưng mà câu a lại  nói tính AD

f) x⁴y³z² : 3xyz²= \(\dfrac{1}{2}\)x³y²

g) (x+y)⁵: (x+y)⁴= (x+y)¹= x+y

h) (x-y)⁷ : (y-x)⁶ = -(y-x)⁷: (y-x)⁶= -(y-x)¹= -(y-x) = -y +x

i) (x-y+z)⁵ : (x-y+z)⁴ = (x-y+z)¹= (x+y-z)= x+y-z

c) 6x³y⁴ : x³y=6y³

d) 30x²y²z: 6xyz

e) 54x⁴y²z: 9x⁴y= 6y³z

a) -28xy⁵z³ : 7xy²z³= -4y³

b) 8x²y²z : 6xyz= 6xy

1 must

2 must / had to

3 has to

4 have to

5 have to

6 must

1 must

2 must / had to

3 has to

4 have to

5 have to

6 must