Solutions of equations 3x-2x = 46, y = 3-5X and 2x + 5Y = 12, 2x + 3Y = 6

46 -227 -1.5 3

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1. Given the absolute value of x-3 + (y + 2 / 5) to the power of 2 = 0, find the value of X of Y
2. If the absolute value of 3-x + the second power of (y + 4) is 0, then the negative y is X=
3. Given the absolute value of x = x + 3, then find the value of (2x + 2) to the power of 2011 Take the answer and send 10
4. What is the equivalent of 2008 * (2007 / 2009),
5. How much is 2008 of 2009 minus 2007 of 2008? The score is fast!
6. How much is 1:1 in 2009 multiplied by 1:1 in 2008 and 2009
7. 2007 in addition to 2007 and 2008 in 2007 multiplied by 2009 + 2008 in 1
8. How to write the formula of 2008 times 2007 of 2009 There are also 4 △ 132 + 6 × 53 (0.86 + 0.86 + 0.86 + 0.86) × 25 261-2.86 × (6.25-641) 261－2.86×（6.25－641） 7.6×21÷［1.9－1.9×（1.9－1.9）］（70＋681）×691 1－1÷2－1÷3－61 19.98×37－199.8×1.9＋1998×0.82
9. 2006 and 2007 divide 2006 by 2006 in a simple way
10. Given | 2a-c + & frac14;; (4b-5a) & sup2;; = 0, find a: B: C
11. If the root of equation 3x-kx-1 = 0 is one third, then K=
12. Given that the equation x ^ 2 + kx-3 = 0 and x ^ 2 + 4x = k-1 have the same root, find K and the same root
13. If one root of the equation 4x ^ - KX + 6 = 0 is 2, then the other root is 2____ ,k=____
14. Given that the equation x ^ 2 + kx-3 = 0 and x ^ 2-4x - (k-1) = 0 have only one equal root, find the value of K and the same root
15. Let x ^ 2 + kx-1 = 0 (1) prove that the equation has two unequal real roots (2) let the two heels of the equation be x1, X2 respectively, Satisfy X1 + x2 = X1 times X2, find the value of K
16. What is the larger area of the figure enclosed by y = | x | and circle x ^ 2 + y ^ 2 = 4
17. The area of the smaller figure enclosed by y = | x | and circle x ^ 2 + y ^ 2 = 4 is a: π / 4 B: π C: 3 π / 4 D: 3 π / 2
18. If it is known that the function f (x) defined on R is an odd function with period 2, then the equation f (x) has at least two properties in the interval [- 2,2]_____ A real number root If f (0) = 0, f (- 1) = - f (1) is obtained by odd function, and if f (2) = f (- 2) = f (0), f (1) = f (- 1) is obtained by period 2, then f (2) = f (- 2) = f (1) = f (- 1) = f (0) = 0, that is, there are at least five real roots I don't understand the answer“ So f (2) = f (- 2) = f (1) = f (- 1) = f (0) = 0 How can f (- 2) be equal to F (1) and f(-1) Where else do you get five real roots?
19. If f (x) is an even function defined on R with period of 3 and f (2) = 0, then the minimum number of solutions of the equation f (x) = 0 in the interval (0,6) is () A. 5B. 4C. 3D. 2
20. It is known that the odd function f (x) defined on R satisfies f (x-4) = - f (x) and is an increasing function in the interval [0,2]. If the equation f (x) = m (M > 0) has four different roots x1, X2, X3, X4 in the interval [- 8,8], then X1 + x2 + X3 + X4 =