Question

Oblicz wartosci liczbowe podanych sum algebriacznych dla x=-1 i y=0,1.
a) 4(5y+1)(1-5y)-5(4x+1)(1-4x)
b)(x+2y)²+(2x-y)²-5(x-y)(x+y)
c)3y(3y-2x)-(3-x)(x+3)-(x-3y)²

Answer 1

Wystarczy jak podstawisz za x, -1 i za y 0,1 czyli
a)4*(5*0,1+1)(1-5*0,1)-5(4*-1+1)(1-4*-1)
4(0,5+1)(1-0,5)-5(-4+1)(1+4)=2+4+4-2+20-5-5-20=-2
i punkt bi c tak samo musisz zrobic

Answer 2

A) 4(5y+1)(1-5y)-5(4x+1)(1-4x)=4(-25y²+1)-5(-16x²+1)=100y²+4+80x²-5=
=100y²+80x²-1=100*(0,1)²+80*(-1)²-1=-100*0,01+80*11=-1+80-1=78

b)(x+2y)²+(2x-y)²-5(x-y)(x+y)=x²+2*x*2y+(2y)²+(2x)²-2*2x*(-y)+y²-5(x²-y²)=
x²+4xy+4y²+4x²-4xy+y²-5x²+5y²=10y²=10*(0,1)²=0,1

c)3y(3y-2x)-(3-x)(x+3)-(x-3y)²=9y²-6xy-9-x²-x²-6xy+9y²=-2x²+18y²-12xy-9=-2*(-1)²+18*(0,1)²-12*(-1)*0,1-9=2+0,18+1,2-9=-5,62

Answer 3

A(2+4)(1+2)(20-5)(1+4)
6*3*15*5
1350
b(-1+0,4)(-1+0,4)+(-2-0.1)(-2-0,11)-5(-1-0.1)=(1-0.4-0.4+0.16)+(4+0,2+0,2-).01)-5(1:0,1-0,1-0,01)=(0.36+4.39)(-5-0.05)=4,75*(-4.95)=23.51