Question

Rozwiąż zadanie 3x+y; 2x-3y; x*y; x/y; y/x; x*2; y*2 (* to mnożenie, *2 to potęga kwadratowa, / to dzielenie/ułamek, # to pierwiastek ) jeżeli :
a) x= (-2 #3) - 8 #5 y= (-4#3) + 3#5
b) x= 2/3 - 5#3 y= 1,5 - #3

Answer 1

A)3x+y
3(-2#3 - 8#5) + (-4#3) +3#5 = -6#3 -24#5 -4#3 +3#5 = -10#3 -21#5

2x-3y
2(-2#3 - 8#5) - 3(-4#3 +3#5) = -4#3 -16#5 +12#3 -9#5 = 8#3 -25#5

x*y
(-2 #3 - 8 #5) * (-4#3 + 3#5) = 8#3 -6#15 +32#15 -24#25 = 8#3 +16#15 -120

x/y
(-2 #3 - 8 #5) / (-4#3 + 3#5)

y/x
(-4#3 + 3#5) / (-2 #3 - 8 #5)

x*2
(-2 #3 - 8 #5)*2 = -1*(2 #3 + 8 #5)*2 = -1*(12 + 320 +32#15) = -332 -32#15

y*2
(-4#3 + 3#5)*2 = 48 +45 - 24#15 = 93 - 24#15

b) 3x+y
3*(2/3 - 5#3) + 1,5 - #3 = 2-15#3 +1,5 - #3 = 3,5 - 14#3

2x-3y
2*(2/3 - 5#3) - 3*(1,5 - #3) = 4/3 - 10#3 - 4,5 +3#3 = 2 5/6 - 7#3

x*y
(2/3 - 5#3) * (1,5 - #3) = 1 -2/3#3 - 7,5#3 +15 = 16 - 8 1/6#3

x/y
(2/3 - 5#3)/(1,5 - #3)

y/x
(1,5 - #3) / (2/3 - 5#3)

x*2
(2/3 - 5#3)*2 = 4/9 +75 -6 2/3#3 = 75 4/9 - 6 2/3#3

y*2
(1,5 - #3)*2 = 2,25 - 3 -3#3 = -0,75 - 3#3