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                Bài làm :

Ta có :

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

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

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

\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)

Thay x+y=5 vào biểu thức trên ; ta được :

\(Q=5^3-2.5^2+3.5+10=125-50+15+10=100\)

Vậy Q=100

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

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

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

\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)

Thay x + y = 5 vào pt ta được :

\(Q=5^3-2.5^2+3.5+10=125-50+15+10=100\)

Vậy Q = 100 <=> x + y = 5

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

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

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

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

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

Với x + y = 5

Q = 5³ - 2.5² + 3.5 + 10

= 125 - 50 + 15 +10

= 100

B=....
<=>B=x^3+y^3+3xy(x+y)-2(x^2+y^2+2xy)+3(x+y)+10
<=>B=(x+y)^3-2(x+y)^2+3(x+y)+10
tại x+y=5 thay vao B ta đc:
B=5^3-2.5^2+3.5+10
B=100

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

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

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

a)  \(A=x^2+2xy+y^2-4x-4y+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1=-2\)

b)  \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+37=100\)

c)  \(C=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2.5+10=25\)

a) \(A=x^2+2xy+y^2-4x-4v+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1=-2\)

Câu 2: 

\(B=x^2+2x+y^2-2x-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2\cdot7+37=49+37+14=100\)

Câu 3: 

\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2\cdot5+10=25\)

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

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

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

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

\(P=3.25-10-100\)

\(P=-35\)

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

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

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

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

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:

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

\(P=5.13-100\)

\(P=-35\)

P = 3x² - 2x + 3y² - 2y + 6xy - 100

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

= 3( x + y )² - 2( x + y ) - 100

Với x + y = 5

=> P = 3.5² - 2.5 - 100 = 75 - 10 - 100 = -35

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

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

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

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

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

Với x + y = 5

=> Q = 5³ - 2.5² + 3.5 + 10 = 100

a. \(P=3x^2-2x+3y^2-2y+6xy-100\)

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

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

\(\Leftrightarrow P=3.5^2-2.5-100\)

\(\Leftrightarrow P=-35\)

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

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

\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)

\(\Leftrightarrow Q=5^3-2.5^2+3.5+10\)

\(\Leftrightarrow Q=100\)

 a) P = 3x² - 2x + 3y² - 2y + 6xy - 100

        =(3x²+3y²+6xy)-(2x+2y)-100

        =3(x²+y²+2xy)-2(x+y)-100

        =(3(x²+2xy+y²-2(x+y)-100

        =3(x+y)²-2(x+y)-100

Thay x+y=5 ta đc:

P=3.5²-2.5-100=-35

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

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

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

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

        =(x+y)³-2(x+y)²+3(x+y)+10

Thay x+y=5 ta đc:

Q=5³-2.5²+3.5+10=100