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= (3x²+3y²⁾ - 2 (x+y) +6xy+1953

A= 3(x²+y²) - 2(x+y) +6xy+1953

A= 3(x²+2xy+y² - 2xy) - 2(x+y) +6xy +1953

A= 3 [(x+y)² -2xy] - 2(x+y) +6xy+1953

A= 3(x+y)² -6xy - 2(x+y) +6xy+1953

A= 3(x+y)² - 2(x+y)+1953

Thayy x+y =5 vào A, ta được: 

A= 3.5² - 2.5 +1953

A= 2018.

A=3(x2+2xy+y2)-2(x+y)-100=3(x+y)2-2.5-100=3.52-110=-35    

B=x3+3x2y+3xy2+y3-2(x2+2xy+y2)+3(x+y)+10=(x+y)3-2(x+y)2+3.5+10=53-2.52+25=100

trả lời:

A=3(x2+2xy+y2)-2(x+y)-100

=3(x+y)2-2.5-100

=3.52-110

=-35

B=x3+3x2y+3xy2+y3-2(x2+2xy+y2)+3(x+y)+10

 =(x+y)3-2(x+y)2+3.5+10

 =53-2.52+25

 =100

học tốt

câu a sai đề, như thế này nè:

A = x³ + y³ - 2x² - 2y² + 3xy (x + y) - 4xy + 3(x + y) +10 

a,A = x³ + y³ - 2x² - 2y² + 3xy (x + y) - 4xy + 3(x + y) +10

A = x³ + y³ - 2x² - 2y² + 3x²y + 3xy² - 4xy + 3x + 3y +10

A = (x³ + 3x²y + 3xy² + y³) - (2x² + 4xy + 2y²) + (3x + 3y) +10

A = (x + y)³ - 2 (x + y)² +3(x + y) +10

Thay x + y = 5 ta được:

A = 5³ - 2. 5² + 3.5 +10 = 100

 b, B = 3x² - 2x + 3y² - 2y + 6xy - 100  

B = (3x² + 3y² + 6xy ) - (2x + 2y) -100

B = 3(x + y)² - 2(x + y) - 100

Thay x + y = 5 ta được:

B = 3.5² - 2.5-100 = -35

xích mích à

tự làm đi đừng ai giúp nhé lần này lại gặp mi nữa rồi

uh đúng đấy

\(A=\frac{1}{3}(3(x+y)-1)^2-\frac{301}{3}\)

a: \(M=x^3+x^2-y^3+y^2+xy-3xy-95\)

\(=\left(x-y\right)^3+\left(x-y\right)^2-95\)

\(=7^3+7^2-95=297\)

b: \(N=3\left[\left(x+y\right)^2-2xy\right]-2\left(x+y\right)+6xy-100\)

\(=3\cdot\left(25-2xy\right)-10+6xy-100\)

=75-6xy-10+6xy-100

=-35

\(a,P=3x^2-2x+3y^2-2y+6xy-100\)

\(P=3\left(x^2+y^2\right)-\left[2\left(x+y\right)\right]+6xy-100\)

\(P=3\left(x^2+y^2+2xy-2xy\right)-2.5+6xy-100\)

\(P=3\left(x+y\right)^2-6xy-10+6xy-100\)

\(P=3.25-10-100\)

\(P=-35\)

\(b,Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(Q=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x^2+y^2+2xy-2xy\right)+3xy.5-4xy+3.5+10\)\(Q=5.\left(x^2+y^2+2xy-3xy\right)-2\left(x+y\right)^2+4xy+15xy-4xy+25\)

\(Q=5.5-15xy-2.25+15xy+25\)

\(Q=25-50+25=0\)

a) P= 3x² -2x + 3y²-2y + 6xy -100

= (3x²+ 3y² + 6xy) - 2(x+y) -100

=3(x² + y² +2xy) - 2(x+y) -100

=3(x+y)² - 2(x+y) -100

=3 . 5² -2 .5 -100

=35

b) Q=x³ + y³ -2x² -2y² + 3xy (x+y) -4xy + 3(x+y) + 10

=(x³ +y³) + 3xy (x+y) + 3(x+y) -4xy -2x² -2y² + 10

=(x+y) (x² -xy +y² ) + 3xy (x+y) + 3 (x+y) - 2 (2xy + x² +y² ) + 10

=(x+y) (x² -xy +y² + 3xy ) + 3(x+y) -2 (2xy + x² + y² ) + 10

=(x+y) (x² +2xy +y² ) + 3(x+y) - 2(x+y)² + 10

= (x+y)³ + 3(x+y) - 2 (x+y)² + 10

=5³ + 3.5 -2. 5²+ 10

=100

a)

\(P=3x^2-2x+3y^2-2y+6xy-100\)

\(P=\left(3x^2+3xy\right)+\left(3y^2+3xy\right)-\left(2x+2y\right)-100\)

\(P=3x\left(x+y\right)+3y\left(x+y\right)-2\left(x+y\right)-100\)

\(P=\left(x+y\right)\left(3x+3y-2\right)-100\)

Vì x + y = 5 (1)

=> 3(x + y) = 15

=> 3x + 3y = 15 (2)

Thay (1), (2) vào P ta được:

\(P=5\left(15-2\right)-100\)

\(P=5.13-100\)

\(P=-35\)