MTC = (x - y)(x² + xy + y²)

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

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

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

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

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

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

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

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

1/x-y-3xy/x^3-y^3+x-y/x^2+xy+y^2

=1/x-y+-3xy/(x-y)(x^2+xy+y^2)+x-y/x^2+xy+y^2

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

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

=2x^2-5xy/(x-y)(x^2+xy+y^2)

\(2x^2y^2\left(-4x^5y^3\right)\)

\(=2\cdot\left(-4\right)\left(x^2\cdot x^5\right)\left(y^2\cdot y^3\right)\)

\(=-8x^7y^5\)

Ta có:

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

Thay \(x=-3\) vào A, ta có:

\(A=2.\left(-3\right)^2=18\)

Vậy A=18

\(A=x\left(x+y\right)-x\left(y-x\right)=x\left(x+y\right)+x\left(x+y\right)=\left(x+y\right).2x=\left(-3+2\right).2.\left(-3\right)=6\)

Ta có:

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

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

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

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

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

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

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

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

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

$(x-1)(x^2+x+1)-x^3-6x=11$

Dùng $(x-1)(x^2+x+1)=x^3-1$:

$x^3-1-x^3-6x=11$

$-6x-1=11$

$-6x=12$

$x=-2$

Vậy $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

Điều kiện: $x\ne-2,-1$.

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

$x^2-1=x^2+4x+4$

$-4x=5$

$x=-\dfrac54$

Vậy $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

Điều kiện: $x\ne-2,-1$.

$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$

$x+1+2x+4=0$

$3x+5=0$

$x=-\dfrac53$

Vậy $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

Điều kiện: $x\ne0,3$.

$\dfrac{9-x^2}{x(x-3)}=1$

$9-x^2=x(x-3)$

$9-x^2=x^2-3x$

$2x^2-3x-9=0$

$(2x+3)(x-3)=0$

$x=-\dfrac32$ hoặc $x=3$

Nhưng $x=3$ không thỏa điều kiện.

Vậy $x=-\dfrac32$.

a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)

b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)

\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)

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

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

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

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

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

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

a) \(18x^4y^3:12\left(-x\right)^3y\)

\(=\left(18:-12\right)\left(x^4:x^3\right)\left(y^3:y\right)\)

\(=-\dfrac{3}{2}xy^2\)

b) \(x^2y^2-2xy^3:\dfrac{1}{2}xy^2\)

\(=\dfrac{xy^2\left(x-2y\right)}{\dfrac{1}{2}xy^2}\)

\(=\dfrac{x-2y}{\dfrac{1}{2}}\)

\(=2x-4y\)