1 if x+1/x=2 then x^2012+1/x^2012=........
2 if A > B > C > D, find the digits A,B,C,D which are from 1 to 9 so that ABCD-DCBA=BDAC
3 solve x+y=28 xy=127
1) is that x + (1/x) or (x+1)/x?
2)
7641 - 1467 = 6174
3)
(x,-y) = (14±√69)
it is x+(1/x)
1. To find the value of x^2012 + 1/x^2012 given x + 1/x = 2, we can use a recursive approach.
First, let's find the value of x^2 + 1/x^2. Squaring the equation x + 1/x = 2 gives us x^2 + 2 + 1/x^2 = 4. Rearranging the terms, we have x^2 + 1/x^2 = 2.
Now, let's square the equation x^2 + 1/x^2 = 2. Squaring both sides gives us (x^2 + 1/x^2)^2 = 4. Expanding the square, we get x^4 + 2 + 1/x^4 = 4.
Continuing this process, we can find x^8 + 2 + 1/x^8 = 4, x^16 + 2 + 1/x^16 = 4, and so on. Eventually, we will reach x^2012 + 2 + 1/x^2012 = 4.
Subtracting 2 from both sides, we get x^2012 + 1/x^2012 = 2.
Therefore, x^2012 + 1/x^2012 = 2.
2. To find the digits A, B, C, D that satisfy the equation ABCD - DCBA = BDAC, we can use trial and error.
Let's start with the highest possible value for A, which is 9. Since A > B > C > D, the maximum value of A is 9. If A = 9, then we have 90CD - DC9 = 9D9C.
From this equation, we can deduce that D must be 1 since we have 90C - C9 = 91C. The maximum value for C is 8 because C > D = 1. If C = 8, we have 908 - 81 = 918, which doesn't satisfy the equation.
Next, let's try A = 8. Using a similar approach, we find that D must be 2 and C must be 7. If C = 7, we have 802 - 28 = 824, which is also not a solution.
Iterating through all possible values, we find that there is no solution where A > B > C > D such that ABCD - DCBA = BDAC.
Therefore, there are no digits A, B, C, D from 1 to 9 that satisfy the equation.
3. To solve the system of equations x + y = 28 and xy = 127, we can use the substitution method.
From the first equation, we can solve for y by subtracting x from both sides: y = 28 - x.
Now we can substitute this value of y into the second equation: x(28 - x) = 127.
Expanding the equation, we have 28x - x^2 = 127.
Rearranging the equation, we get x^2 - 28x + 127 = 0.
Now we can either factor the quadratic or use the quadratic formula to find the solutions for x.
Factoring, we have (x - 7)(x - 18) = 0.
So the possible values for x are 7 and 18.
Substituting these values back into the equation y = 28 - x, we find that when x = 7, y = 21 and when x = 18, y = 10.
Therefore, the solutions to the system of equations are (x, y) = (7, 21) and (18, 10).