\(a,P=3x^2-2x+3y^2-2y+6xy-100\)
\(P=3\left(x^2+y^2\right)-\left[2\left(x+y\right)\right]+6xy-100\)
\(P=3\left(x^2+y^2+2xy-2xy\right)-2.5+6xy-100\)
\(P=3\left(x+y\right)^2-6xy-10+6xy-100\)
\(P=3.25-10-100\)
\(P=-35\)
\(b,Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(Q=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x^2+y^2+2xy-2xy\right)+3xy.5-4xy+3.5+10\)\(Q=5.\left(x^2+y^2+2xy-3xy\right)-2\left(x+y\right)^2+4xy+15xy-4xy+25\)
\(Q=5.5-15xy-2.25+15xy+25\)
\(Q=25-50+25=0\)
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:
\(P=5\left(15-2\right)-100\)
\(P=5.13-100\)
\(P=-35\)