c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)

\(=x^2+x-10\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`5x^3 - x - 1/2`

Đã thu gọn?

`b)`

`(3xy - x^2 + y) * 2/3x^2y`

`= 3xy * 2/3 x^2y - x^2* 2/3x^2y + y*2/3x^2y`

`= 2x^3y^2 - 2/3x^4y + 2/3x^2y^2`

`c)`

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

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

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

`d)`

`(x^2 - 2x +1) (x-1)`

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

`= x^3 - x^2 - 2x^2 + 2x + x -1`

`= x^3 -3x^2 + 3x - 1`

d: =(x-1)^3=x^3-3x^2+3x-1

c: =-2x^4y+5/2x^2y^2-x^2y

\(a.2x\left(x-1\right)-3\left(x^2+4x\right)+x\left(x+2\right)\) 

\(=2x^2-2x-3x^2-12x+x^2+2x\) 

\(=-12x\) 

\(b.\left(2x-3\right)\left(3x+5\right)-\left(x-1\right)\left(6x+2\right)+3-5x\) 

\(=6x+10x-9x^2-15-6x^2-2x-6x-2+3-5x\) 

\(=-15x^2+3x-14\) 

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

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

\(=y^3+x^2y\) 

a) \(3\left(2x-3\right)+5\left(x+2\right)=6x-9+5x+10=11x+1\)

b) \(3x\left(2x-8\right)+\left(6x+2\right)\left(5-x\right)=6x^2-24x+30x-6x^2+10-2x=4x+10\)

c) \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2=x^2-9-x^2+10x-25=10x-34\)

d) \(\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-3x^2y+3xy^2-y^3-x^3+y^3=3xy^2-3x^2y\)

Tk :

a: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

b: Ta có: \(3x\left(2x-8\right)-\left(6x-2\right)\left(x-5\right)\)

\(=6x^2-24x-6x^2+30x+2x-10\)

\(=8x-10\)

c: Ta có: \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2\)

\(=x^2-9-x^2+10x-25\)

\(=10x-34\)

Lời giải:

a.

$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$

$=4x^2-12x+9+4x^2-4x-15$

$=24-8x$
b.

$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$

c.

$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$

$=6x^2-24x-6x^2-32x+10$

$=8x-10$

d.

$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$

$=x^2-9-x^2+10x-25=10x-34$

e.

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

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

a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)

\(=4x^2-12x+9+2x-4x^2+15-6x\)

\(=-16x+24\)

b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)

\(=6x^2-24x+30x-6x^2-10+2x\)

\(=8x-10\)

Ta có : \(x^2+2x+y^2-2y-2xy+65\)

\(=\left(x-y\right)^2+2\left(x-y\right)+65\)

Mà \(x=y+5\)

\(\Rightarrow x-y=5\)

- Thay x  - y = 5 vào đa thức trên ta được :

\(=\left(x-y\right)^2+2\left(x-y\right)+65=100\)

Vậy ...

a: \(\Leftrightarrow\left(2x+1\right)^3=8\cdot25-75=125\)

=>2x+1=5

hay x=2

c: x=2; y=0

a: \(=15x^5y^3-6x^4y^2-6x^3y^3\)

c: \(=2x^4-2x^2-3x^3+3x+x^2-1\)

\(=2x^4-3x^3-x^2+3x-1\)