\(x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=100\)

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:

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

\(P=5.13-100\)

\(P=-35\)

Đáp án đúng : B

Chắc đề bài là \(Q=\dfrac{3}{9x^2+6xy+y^2}+\dfrac{3}{3x^2+6xy+2y^2}\)

Từ giả thiết ta có:

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

\(\Leftrightarrow2x\left(x^2+y^2\right)+y\left(x^2+y^2\right)=2\left(x^2+y^2\right)\)

\(\Leftrightarrow2x+y=2\)

Do đó:

\(Q=3\left(\dfrac{1}{9x^2+6xy+y^2}+\dfrac{1}{3x^2+6xy+2y^2}\right)\)

\(Q\ge\dfrac{3.4}{12x^2+12xy+3y^2}=\dfrac{4}{\left(2x+y\right)^2}=1\)

\(Q_{min}=1\) khi \(\left\{{}\begin{matrix}2x+y=2\\9x^2+6xy+y^2=3x^2+6xy+2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{6}-2\\y=6-2\sqrt{6}\end{matrix}\right.\)

3x^2+3y^2+4xy-2x+2y+2=0

=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0

=>x=1 và y=-1

M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

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