3+2*A+B*4/2-4 = 3 + 2A + 2B - 4 = 2A + 2B - 1
3 + 5 = 4*b so 8 = 4*b Divide both sides by 4: 2 = b
(a + 2 - b) (a + 2 + b)
10a + 5b
in ones digit arithmetic to define a+b you take the integer a+b and drop all but the one's digit (same for a x b). 1+2+3+4+5+6+7=8 because you drop all but the one's digit.
3+2*A+B*4/2-4 = 3 + 2A + 2B - 4 = 2A + 2B - 1
b^2 - 7b + 12 = b^2 - 4b - 3b + 12 = b(b -4) -3(b - 4) = (b - 3)(b - 4)
a+b+2+3=5+a+b
3 + 5 = 4*b so 8 = 4*b Divide both sides by 4: 2 = b
(b + 2)(b + 2) or (b + 2)2
(a + 2 - b) (a + 2 + b)
10a + 5b
in ones digit arithmetic to define a+b you take the integer a+b and drop all but the one's digit (same for a x b). 1+2+3+4+5+6+7=8 because you drop all but the one's digit.
3b+4 = b+12 3b-b = 12-4 2b = 8 b = 4
1 ------ a+b=4 2 ------ ab=2 ====> 3. b = 2/a Sub 3 into 1 ===> a + 2/a = 4 mutiply both sides by a ===> a2 +2 -4a = 0 use quadratic formula to find a ==> a= 2 +sqrt(2) or a' = 2- sqrt(2) use these two values of a to find a value for b using equation 3 ===> using a, b= 2-sqrt(2) and using a', b= 2+sqrt(2) hence a4 + b4 = (2+sqrt(2))4 + (2-sqrt(2))4 = 136 (for both values of a and b)
4/b-3 + 3/b = -2b/b-3 b = 1 or b = -4.5
No. Here is a proof by counterexample that it does not.Given ab + bc + ca = 3:Assume toward a contradiction that abc is a cube. Then a = b = c.Without loss of generality, let a = 2, b = 2, and c = 2.Then ab = 4, bc = 4, and ca = 4.ab + bc + ca = 4 + 4 + 4 = 12.Therefore, 12 = 3, which is false, and so the original statement is false.