1: \(f\left(-x\right)=\left(-x\right)^2=x^2\)

Vậy: Hàm số này chẵn

y '   =   x 2   +   x   -   2

a) x = {-2; 1}

b)x = { -1; 0}

c) x = {-4; 3}

\(\begin{array}{l}T + H = 3{x^2}y - 2x{y^2} + xy + \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy - 2{x^2}y + 3x{y^2} + 1\\ = \left( {3{x^2}y - 2{x^2}y} \right) + \left( { - 2x{y^2} + 3x{y^2}} \right) + xy + 1\\ = {x^2}y + x{y^2} + xy + 1\\T - H = 3{x^2}y - 2x{y^2} + xy - \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy + 2{x^2}y - 3x{y^2} - 1\\ = \left( {3{x^2}y + 2{x^2}y} \right) + \left( { - 2x{y^2} - 3x{y^2}} \right) + xy - 1\\ = 5{x^2}y - 5x{y^2} + xy - 1\end{array}\)

Chọn B.

\(A,VT=x^3+y^3+x^3-y^3=2x^3=VP\\ B,VT=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left(x^2+2xy+y^2-xy\right)\\ =\left(x-y\right)\left[\left(x+y\right)^2-xy\right]=VP\)

Sửa câu b \(cm:x^3-y^3=\left(x-y\right)\left[\left(x+y\right)^2-xy\right]\)

g: (x+3y)(x-3y+2)

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

=x^2-9y^2+2x+6y

h: (x+2y)(x-2y+3)

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

=x^2-4y^2+3x+6y

i: (x^2-xy+y^2)(x+y)

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

=x^3+y^3

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

k: (5x-2y)(x^2-xy-1)

=5x*x^2-5x*xy-5x-2y*x^2+2y*xy+2y

=5x^3-5x^2y-5x-2x^2y+2xy^2+2y

=5x^3-7x^2y+2xy^2-5x+2y

l: (x^2y^2-xy+y)(x-y)

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

Ta có: \(\dfrac{x^2+xy}{x^2+xy+y^2}-\left(\dfrac{x\left(2x^2+xy-y^2\right)}{x^3-y^3}-2+\dfrac{y}{y-x}\right):\dfrac{x-y}{x}-\dfrac{x}{x-y}\)

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

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

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

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

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

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

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

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

Lời giải:

a. $xy(x+y)-y(x+y)^2+y^2(x-y)$

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

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

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

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

$=y^2(-2y)=-2y^3$

b.

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

$=[x(x+y)^2-y(x+y)^2]-(x^2-xy)$

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

$=(x-y)[(x+y)^2-x]=(x-y)(x^2+2xy+y^2-x)$

a: \(xy\left(x+y\right)-y\left(x+y\right)^2+y^2\left(x-y\right)\)

\(=\left(x+y\right)\left[xy-y\left(x+y\right)\right]+y^2\left(x-y\right)\)

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

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

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

b: \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)

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

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

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