a: A=2/3x^2y+4x^2y=14/3x^2y

=14/3*9*7=294

b: B=xy^2(1/2+1/3+1/6)=xy^2=3/4*1/4=3/16

c: C=x^3y^3(2+10-20)=-8x^3y^3

=-8*1^3(-1)^3=8

d: D=xy^2(2018+16-2016)

=18xy^2

=18(-2)*1/9=-4

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

b)\(\left(x-y\right)^5:\left(y-x\right)^4=\left(x-y\right)^5:\left(x-y\right)^4=x-y\)

c)\(\left(5x^4-3x^3+x^2\right):3x^2=\frac{5}{3}x^2-x+\frac{1}{3}^{ }\)

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

a: \(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{4}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)

\(=-\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{-1}{18}\)

\(=\dfrac{-1}{72}\)

b: \(B=\left(-1\right)^2\cdot3^2+\left(-1\right)\cdot3+\left(-1\right)^3+3^3\)

\(=9-3-1+27=36-4=32\)

f: \(=x^2-9-x^2+6x-9=6x-18\)

a, \(x^2\) + 6x + 5 = 0
=>\(x^2\) + x + 5x +5 = 0
=>x(x + 1) + 5(x + 1) = 0
=>(x + 1)(x + 5) = 0
=> x + 1 =0 hoặc x + 5 =0
=> x = -1 hoặc x = -5

c) \(\dfrac{x+3}{x-1}+\dfrac{2x+5}{x-1}+\dfrac{14-3x}{1-x}\)

\(=\dfrac{x+3}{x-1}+\dfrac{2x+5}{x-1}-\dfrac{14-3x}{x-1}\)

\(=\dfrac{x+3+2x+5-14+3x}{x-1}\)

\(=\dfrac{6x-6}{x-1}\)

\(=\dfrac{6\left(x-1\right)}{x-1}\)

\(=6.\)

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

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

\(=-\dfrac{3x\left(x+y\right)+6y\left(x-y\right)+3y\left(3y-x\right)}{2\left(x-y\right)\left(x+y\right)}\)

\(=-\dfrac{3x^2+3xy+6xy-6y^2+9y^2-3xy}{2\left(x-y\right)\left(x+y\right)}\)

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

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

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

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

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

\(\hept{\begin{cases}3x=2y\\2x+y=3\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{3}{2}.x\\2x+\frac{3}{2}.x=3\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{3}{2}.x\\\frac{7}{2}.x=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{6}{7}\\y=\frac{9}{7}\end{cases}}}\)

\(\hept{\begin{cases}\frac{x}{3}=\frac{3y}{4}\\3x-y=4\end{cases}\Leftrightarrow\hept{\begin{cases}4x=9y\\3x-y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9y}{4}\\\frac{3.9}{4}y-y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{4}.y\\\frac{23}{4}.y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{4}.y\\y=\frac{16}{23}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{36}{23}\\y=\frac{16}{23}\end{cases}}}\)

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