Odpowiedzi

2009-11-11T21:28:42+01:00
A) dla x=3; y=-0,25
(2x-y)(x+2y)+(3y-x)(x-2y)=2x²+4xy-xy-2y²+3xy-6y²-x²+2xy=x²+8xy-8y²=3²+8*3*(-0.25)-8(-0,25)²=9-6-8*0.0625=9-6-0,5=2,5

c) dla x=2,5 y=2
[(x+y)²-(x-y)²]:(4xy)=[x²+2xy+y²-x²+2xy-y²]:4xy=4xy:4xy=4*2,5*2:4*2,5*2=
=20:20=1

d) dla x= √5 y=√3
(x-2y+1)(2x+y-1)+(x-1)(3y-1)=2x²+xy-x-4xy-2y²+2y+2x+y-1+3xy-x-3y+1=2x²-2y²=
=2(√5)²-2(√3)²=2*5-2*3=10-6=4
2009-11-11T21:29:56+01:00
X=2,5 y=2
[(2,5+2)²-(2,5-2)²]÷(4*2,5*2)=[20,25-0,25]÷20=20÷20=1
x=√5 y=√3
(2x²+xy-x-4xy-2y²+2y+2x+y-1)+(3xy-x-3y+1)=2x²-2y²
2*(√5)²-2*(√3)²=2*5-2*3=10-6=4