`a, (4x^3y^2 - 8x^2y + 10xy) : 2xy`

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

`b, 7x^4y^2 - 2x^2y^2 - 5x^3y^4 : 3x^2y`

`= 7/3 x^2y - 3/2y - 5/3xy^3`

Chọn B nhé bạn

a: =x^2(x^2+2x+1)

=x^2(x+1)^2

b: =x^3+3x^2y+3xy^2+y^3-x-y

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

=(x+y)[(x+y)^2-1]

=(x+y)(x+y-1)(x+y+1)

c: =5(x^2-2xy+y^2-4z^2)

=5(x-y-2z)(x-y+2z)

1. Ta có: $x+y+4=0 \Rightarrow x+y=-4$.

Xét: $A=2(x^3+y^3)+3(x^2+y^2)+10xy$.

Ta có: $x^3+y^3=(x+y)^3-3xy(x+y)$ nên: $x^3+y^3=(-4)^3-3xy(-4)=-64+12xy$.

Lại có: $x^2+y^2=(x+y)^2-2xy=16-2xy$.

Thay vào biểu thức $A$:

$A=2(-64+12xy)+3(16-2xy)+10xy$

$=-128+24xy+48-6xy+10xy$

$=-80+28xy$.

Ta có: $(x-y)^2\ge0$

$\Rightarrow (x+y)^2-4xy\ge0$

$\Rightarrow 16-4xy\ge0$

$\Rightarrow xy\le4$.

=> $A=-80+28xy\le-80+28\cdot4=32$.

Dấu “=” xảy ra khi: $x=y=-2$.

Vậy: $\boxed{A_{max}=32}$.

2. Đặt: $t=xy$.

Ta có: $x^4+y^4=(x^2+y^2)^2-2x^2y^2$.

Mà: $x^2+y^2\ge2xy=2t$ nên: $x^4+y^4\ge(2t)^2-2t^2=2t^2$.

Theo giả thiết: $x^4+y^4-7=xy(3-2xy)$

$\Rightarrow x^4+y^4-7=t(3-2t)$.

Do đó: $2t^2-7\le3t-2t^2$

$\Rightarrow 4t^2-3t-7\le0$.

Giải bất phương trình:

$4t^2-3t-7=0$

$\Rightarrow \Delta =(-3)^2-4\cdot4\cdot(-7)=121$

$\Rightarrow \sqrt\Delta=11$.

Suy ra: $t=\dfrac{3\pm11}{8}$

$\Rightarrow t=-1$ hoặc $t=\dfrac74$.

Vì: $4t^2-3t-7\le0$ nên: $-1\le t\le\dfrac74$.

Vậy: $\boxed{M_{min}=-1}$.

a,\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)

\(=\dfrac{2y}{3\left(x+y\right)^2}\)

b,\(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

\(=\dfrac{\left(x^2-x\right)+\left(-xy+y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)

\(=\dfrac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)

\(=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)

\(=\dfrac{x-y}{x+y}\)

c,\(\dfrac{3x^2-12x+12}{x^4-8x}\)

\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)

\(=\dfrac{3\left(x-2\right)^2}{x\left[\left(x-2\right)\left(x^2+2x+4\right)\right]}\)

\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)