a: \(A=\dfrac{3}{2}x^2+6x+3\)

b: \(B=5xy-\dfrac{2}{3}xy+\dfrac{1}{2}x^2y+2x^2y=\dfrac{5}{2}x^2y+\dfrac{13}{3}xy\)

a) \(2x^2+x-\dfrac{1}{2}x^2+5x+3\)\(\)

= \(\left(2x-\dfrac{1}{2}x^2\right)+\left(x+5x\right)+3\)

= \(\dfrac{3}{2}x^2+6x+3\)

Vậy A = \(\dfrac{3}{2}x^2+6x+3\)

\(=12ab^2-3.2b^3-6ab^2+3b^3-1\)

\(=6ab^2-0.2b^3-1\)

a, `(8x^3-4x^2): 4x -(4x^2-5x) : 2x + (2x)^2`

`=4x (2x^2-x) : 4x - 2x(2x-5/2 ) :2x + 4x^2`

`=2x^2-x-2x+5/2+4x^2`

`=6x^2-3x+5/2`

b, `(3x^3-x^2y) :x^2 -(xy^2+x^2y) :xy + 2x(x+1)`

`=x^2 (3x-y) :x^2 -xy(y+x) + (2x^2+2x)`

`=3x-y-y-x+2x^2+2x`

`=2x^2+4x-2y`

a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)

\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)

\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)

\(=2a^2-b^2\)

b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)

\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)

\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)

\(=-7ab+b^2\)

c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)

\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)

\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)

\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)

\(=-7bx+3b^2+2x^2\)

d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)

\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)

\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)

\(=-5ax+32x^2-30a^2\)

a: =2ab+8a^2-b^2-4ab+2ab-6a^2

=2a^2-b^2

b: =6a^2-9ab-4ab+6b^2-6a^2+6ab

=-7ab+6b^2

c: =10bx-5b^2-16bx+8b^2+2x^2-xb

=3b^2+2x^2-7xb

d: =2xa+30x^2+5ax+2x^2-30a^2-12ax

=32x^2-30a^2-5ax

Điều kiện x∈N, x≥3

Vậy x = 4

Nhận xét. Học sinh có thể quên đặt điều kiện cho x nên tìm ra 3 nghiệm( x=0,x=-1,x=4)

Chọn B

Chọn D

Ta có: C = 2A + 3B = 2(x2 - x2y + 5y2+ 5) + 3(3x2+ 3xy2 - 2y2 - 8)

= (2x2 - 2x2y + 10y2+ 10) + (9x2+ 9xy2- 6y2 - 24)

= 11x2 - 2x2y + 9xy2+ 4y2- 14

`#3107`

`a)`

`A=`\(3x^4 + \dfrac{1}3xyz - 3x^4 - \dfrac{4}3xyz + 2x^2y - 6z\)

`= (3x^4 - 3x^4) + (1/3xyz - 4/3xyz) + 2x^2y - 6z`

`= -xyz + 2x^2y - 6z`

Thay `x = 1; y = 3` và `z = 1/3` vào A

`A = -1*3*1/3 + 2*1^2*3 - 6*1/3`

`= -1 + 6 - 2`

`= 6 - 3`

`= 3`

Vậy, `A=3`

`b)`

`B=`\(4x^3 - \dfrac{2}7xyz - 4x^3 - \dfrac{4}3xyz + 4x^2y\)

`= (4x^3 - 4x^3) + (-2/7xyz - 4/3xyz) + 4x^2y`

`= -34/21 xyz + 4x^2y`

Thay `x = -1; y = 2` và `z = -1/2` vào B

`B = -34/21*(-1)*2*(-1/2) + 4*(-1)^2 * 2`

`= -34/21 + 8`

`= 134/21`

Vậy, `B = 134/21`

`c)`

`C=`\(4x^2 + \dfrac{1}2xyz - \dfrac{2}3xy^2z - 5x^2yz + \dfrac{3}4xyz\)

`= 4x^2 + (1/2xyz + 3/4xyz) - 2/3xy^2z - 5x^2yz `

`= 4x^2 + 5/4xyz - 2/3xy^2z - 5x^2yz`

Ta có:

`|y| = 2`

`=> y = +-2`

Thay `x = -1; y = 2` và `z = 1/2` vào C

`4*(-1)^2 + 5/4*(-1)*2*1/2 - 2/3*(-1)*2^2*1/2 - 5*(-1)^2*2*1/2`

`= 4 - 5/4 + 4/3 - 5`

`= -11/12`

Vậy, với `x = -1; y = 2; z = 1/2` thì `B = -11/12`

Thay `x = -1; y = -2; z = 1/2`

`B = 4*(-1)^2 + 5/4*(-1)*(-2)*1/2 - 2/3*(-1)*(-2)^2*1/2 - 5*(-1)^2*(-2)*1/2`

`= 4 + 5/4 + 4/3 + 5`

`= 139/12`

Vậy, với `x = -1; y = -2; z = 1/2` thì `B = 139/12.`