Bài 2: Nhân đa thức với đa thức

\(=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}\\ =\dfrac{5}{3}\)

Cậu cần bài nào nhỉ ? Cần hết thì tách ra từng bài thôi nha cậuu

1:

a: =x^3-6x^2+12x-8

b: (2x-3y)^3

=8x^3-36x^2y+54xy^2-27y^3

c: (x+y/x)^3

=x^3+3xy+3y^2/x+y^3/x^3

2:

a: =(x-3)^3

b: =(-1/2x+1)^3

c: =(x^2-1/2y)^3

a) 6x² + 3xy² + A = x² + y² - 2xy²

A = x² + y² - 2xy² - 6x² - 3xy²

A = (x² - 6x²) + (-2xy² - 3xy²) + y²

A = -5x² - 5xy² + y²

b) B - (2xy - 4y²) = 5xy + x² - 7y²

B = 5xy + x² - 7y² + 2xy - 4y²

B = (5xy + 2xy) + (-7y² - 4y²) + x²

B = 7xy - 11y² + x²

A=6x^3-2x^2+2x-6x^3+3x^2-x^2-3x

=-x

=-2023

`A=2x(3x^2-x+1)-3x^2(2x-1)-x(x+3)`

`= 6x^3 -2x^2 +2x-(6x^3 -3x^2)-(x^2+3x)`

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

`=(6x^3 -6x^3 )+(-2x^2 +3x^2-x^2)+(2x-3x)`

`= -x`

Thay `x=2023` vào `-x`

Ta có : `-x=-2023`

\(a,\left(1-2x\right)\left(x+3\right)+\left(x+1\right)\left(2x-1\right)=14\\ \Leftrightarrow x-2x^2+3-6x+2x^2+2x-x-1=14\\ \Leftrightarrow-2x^2+2x^2+x-6x+2x-x=14+1-3\\ \Leftrightarrow-4x=12\\ \Leftrightarrow x=-3\)

\(b,\left(3x^2+x+2\right)-\left(2x+1\right)\left(2+x\right)-\left(x+4\right)\left(x-5\right)=5\\ \Leftrightarrow\left(3x^2+x+2\right)-4x-2x^2-2-x-x^2-4x+5x+20=5\\ \Leftrightarrow3x^2-2x^2-x^2+x-4x-x-4x+5x=5-20+2-2\\ \Leftrightarrow-3x=-15\\ \Leftrightarrow x=15\)

\(c,3x^2+4\left(x-1\right)\left(x+1\right)-7x\left(x-1\right)=x+12\\ \Leftrightarrow3x^2+4\left(x^2-1\right)-7x^2+7x-x=12\\ \Leftrightarrow3x^2+4x^2-7x^2+7x-x=12+4\\ \Leftrightarrow6x=16\\ \Leftrightarrow x=\dfrac{8}{3}\)

\(a,\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\\ =x^2.\dfrac{1}{2}x-5x^2-2x.\dfrac{1}{2}x+2x.5+3.\dfrac{1}{2}x-15\\ =\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\\ =\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

\(b,\left(x^2y^2-\dfrac{1}{3}xy+2y\right)\left(x-2y\right)\\ =x^3y-2x^2y^3-\dfrac{1}{3}x^2y+\dfrac{2}{3}xy^2+2xy-4y^2\)

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

\(=\dfrac{1}{2}x^3-5x^2-x+10x+\dfrac{3}{2}x-15\)

\(=\dfrac{1}{2}x^3-5x^2+\dfrac{48}{5}x-15\)

b) \(\left(x^2y^2-\dfrac{1}{3}xy+2y\right)\left(x-2y\right)\)

\(=x^3y^2-2x^2y^3-\dfrac{1}{3}x^2y+\dfrac{2}{3}xy^2+2xy-4y^2\)

=-2x^2-3x-10x-15-2x^2+6x+x+7

=-4x^2-6x-8

\(=-\left(x+5\right)\left(2x+3\right)-2x^2+6x+x+7\\ =-\left(2x^2+3x+10x+15\right)-2x^2+7x+7\\ =-2x^2-3x-10x-15-2x^2+7x+7\\ =-4x^2-6x-22\)

=2x^2+3x-10x-15-2x^2+6x+x+7

=-8

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1