a) (x + 3y) (2x²y - 6xy²)

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

= 2xy [(x + 3y) (x - 3y)]

= 2xy (x² - 3y²)

b) (6x⁵y² - 9x⁴y³ + 15x³y⁴) : 3x³y²

= (6x⁵y² : 3x³y²) + (-9x⁴y³ : 3x³y²) + (15x³y⁴ : 3x³y²)

= [(6 : 3) (x⁵ : x³) (y² : y²)] + [(-9 : 3) (x⁴ : x³) (y³ : y²)] + [(15 : 3) (x³ : x³) (y⁴ : y²)]

= 2x² + (-3xy) + 5y²

= 2x² - 3xy + 5y²

#Học tốt!!!

a) \(\left(x+3y\right)\left(2x^2y-6xy^2\right)\)

\(=x\left(2x^2y-6xy^2\right)+3y\left(2x^2y-6xy^2\right)\)

\(=2x^3y-6x^2y^2+6x^2y^2-18xy^3\)

\(=2x^3y-18xy^3\)

b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)

\(=6x^5y^2:3x^3y^2-9x^4y^3:3x^3y^2+15x^3y^4:3x^3y^2\)

\(=2x^2-3xy+5y^2\)

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

\(=\left(2x+3-2x-5\right)^2\)

\(=\left(-2\right)^2=4\)

d) \(\left(y+3\right)^3-\left(3-y\right)^2-54y\)

\(=y^3+9y^2+27y+27-\left(x^2-6x+9\right)-54y\)

\(=y^3+9y^2-27y+27-x^2+6y-9\)

\(=y^3+9y^2-x^2-21y+18\)

\(\dfrac{3}{2x^2+y}+\dfrac{5}{xy^2+}+\dfrac{x}{y^3}\)

=\(\dfrac{3xy^5}{xy^2.y^3\left(2x^2+y\right)+}+\dfrac{10y^3x^2+5y^4}{xy^2.y^3\left(2x^2+y\right)}+\dfrac{2x^4y^2+x^2y^3}{xy^2.y^3\left(2x^2+y\right)}\)

=\(\dfrac{3xy^5+10y^3x^2+5y^4+2x^4y^2+x^2y^3}{xy^5\left(2x^2+y\right)}\)

=\(\dfrac{3xy^5+11y^3x^2+5y^4+2x^4y^2}{xy^5\left(2x^2+y\right)}\)

   ủa đáp án cứ sao sao:<

a)

b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)

\(=2x^2-3xy+5y^2.\)

c)

Chúc bạn học tốt!

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)

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) \(x^2y^2-2xy^3:\dfrac{1}{2}xy^2\)

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

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

\(=2x-4y\)

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 $x=-\dfrac32$.

Ta có:

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

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

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

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

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

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

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

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