P = 3x² - 2x + 3y² - 2y + 6xy - 100

= (3x² + 6xy + 3y²) - (2x + 2y) - 100

= 3(x² + 2xy + y²) - 2(x + y) - 100

= 3(x + y)² - 2.5 - 100

= 3. 5² -10 - 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 + 3.5 + 10

= (x³ + 3x²y + 3xy² + y³) - (2x² + 4xy + 2y²) + 15 + 10

= (x + y)³ - 2(x² + 2xy + y²) + 25

= 5³ - 2(x + y)² +25

= 125 - 2. 5² + 25

= 125 - 50 + 25 = 100

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

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

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

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

\(P=3.25-10-100\)

\(P=-35\)

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

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

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

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

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:

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

\(P=5.13-100\)

\(P=-35\)

Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

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

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

b)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

(l ike nha)

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