If (M-3) x ^ | m | - 5-4m = 0 is a one variable linear equation about X, find the value of m ^ - 2m + 1 / 2 of M

If (M-3) x ^ | m | - 5-4m = 0 is a one variable linear equation about X, find the value of m ^ - 2m + 1 / 2 of M

The solution is m = + 1 or - 1 (1)
Bring (1) into m ^ - 2m + 1 / M
You can get the answer
Let u = R, set a = {x  2 + 3x + 2 = 0}, B = {x  2 + (M + 1) x + M = 0}. If B is contained in a, the value of M is obtained
When B = {- 1}, △ = (m-1) & # 178; = 0, # M = 1, then substituting x = - 1 into m can be any value,  M = 1 holds, when B = {- 2}, △ = (m-1) & # 178; = 0,  M = 1, then substituting x = - 2 into M = 2,  does not hold, when B = {- 1, - 2},  M = (m-1) & # 178; >
Let u = R, a = {x} | x ^ 2 + 3x + 2 = 0}, B = {x} x ^ 2 | + (M + 1) x + M = 0}, if the intersection B of (A's complement) is not equal to the empty set, find the value of real number M
The complement of a = {- 2, - 1} is a number divided by - 2. - 1
So (- M-1 plus minus radical (M + 1) ^ 2 / 2 is not equal to - 2. - 1
The results of their own calculation, can cultivate your ability
I can only help you here
When the 12th power of 2 is equal to the 4th power of a and the 4th power of 64 is equal to the B power, find the square of (4 / a plus 8 / b) (4 / a minus 8 / b) plus (4 / a plus 8 / b)
2 ^ 12 = a ^ 4 = 64 ^ B means a ^ 4 = (2 ^ 3) ^ 4 = 8 ^ 4, so a = 864 ^ B = (2 ^ 6) ^ 2 = 64 ^ 2, so B = 2 (A / 4 + B / 8) * (A / 4-b / 8) + (A / 4 + B / 8) ^ 2 = (A / 4 + B / 8) (A / 4-b / 8 + A / 4 + B / 8) = (A / 4 + B / 8) * A / 2 = (8 / 4 + 2 / 8) * 8 / 2 = (2 + 1 / 4) * 4 = 9
Eight
Several factorization: (hope speed, hurry!)
1. Given X-Y = 2, the square of X-Y = 6, then xy = ()
2. On a big circle with radius r, dig out 9 small round holes with radius R. when r = 70cm and R = 10cm, the area of the remaining part is () square centimeter. (PAI = 3.14)
3. Solve the equation: the third power of x-9x = the second power of X-9
(there should be a process, thank you!)
X ^ 2-y ^ 2 = 6, i.e. (x + y) (X-Y) = 6, X-Y = 2, so x + y = 3. Solving the equation, we get x = 5 / 2, y = 1 / 2, so xy = 5 / 4 = 1.252. Area formula s = Pai × R ^ 2. We can deduce large circle area s = Pai × 70 ^ 2, and subtract 9 small original areas, that is 9s small = 9 × Pai × 10 ^ 2
1. X & # 178; - Y & # 178; = (x + y) (X-Y), X-Y = 2, so x + y = 3;
(x + y) ² - (X-Y) ² = 4xy = 5, so xy = 5 / 4
2、S=πR²-πr²=π(R+r)(R-r)=3.14*80*60=15072m²
3. X & # 179; - 9x = x & # 178; - 9 = > x (X & # 178; - 9) = x & # 178; - 9, so x = 1 or X & # 178; = 9, that is, x = ± 3
The square of X-Y = (X-Y) (x + y) = 6, so x + y = 3. From x + y = 3, X-Y = 2, x = 5 / 2, y = 1 / 2, xy = 5 / 4
70x70x3.14-9x10x10x3.14=12560
The third power of X - 9x = the second power of X - 9
The third power of X - the second power of X - 9x + 9 = 0
(x-1)(x-3)(x+3)=0
x=1,x=3,x=-3
It is known that the power of N + 11 of 3 is divisible by 10. It is proved that the power of N + 4 of 3 and the power of M + 2 of 11 can also be divisible by 10
Let 3 ^ n + 11 ^ m = 10K (k is a positive integer), then 3 ^ n = 10k-11 ^ m3 ^ (n + 4) + 11 ^ (M + 2) = 81 (10k-11 ^ m) + 121 * 11 ^ m = 510k + (121-81) * 11 ^ m = 510k + 40 * 11 ^ m = 10 [51K + 4 * 11 ^ m] K and m are all positive integers, and 〈 3 ^ (n + 4) + 11 ^ (M + 2) is divisible by 10
Several problems of factorization
(a-b)3+(a+b)3-8a3
(x+y+z)3+93x-2y-3z)3-(4x-y-2z)3
a4+b4+c4-2a2b2-2b2c2-2c2a2
It is known that the m-th power of 3 + 11 can be divisible by 10 and the n-th power of 3 is an integer. It is proved that the m-th power of 3 + 4 plus 11 can also be divisible by 10
Let n times of three plus m times of 11 be 10K, and then decompose the proved formula. The proved formula is 10K + 80 * 3N times + 120 * 11m times = 10p, and P is a natural number
Some questions about factorization
1. Given a + B = 2 / 3, ab = 2, find the value of the algebraic formula A & sup2; B + 2A & sup2; B & sup2; + AB & sup2
2. Prove that. N5-5n & sup3; + 4N can be divisible by 120 when n is an integer greater than 2
3. Known a + B + C = 0. Prove: A & sup3; + A & sup2; C + B & sup2; c-abc + B & sup3; = 0
First: the original formula = AB (a + 2Ab + b) = 2 (2 / 3 + 2 × 2)
The second way: the original formula = n (n2-4) (n2-1). Whenever n takes any integer greater than 2, there is a factor of 120, so it can be divided by 120
Third: prove: (A3 + B3) + C (A2 + B2 AB) = (a + b) (A2 + B2 AB) + C (A2 + B2 AB) = (A2 + B2 AB) (a + B + C) because a + B + C = 0, so the proof is true
m. N is a positive integer. If M is greater than N, we prove that the n-th power of 2 minus 1 can divide the m-th power of 2 minus 1
The following description is more complicated, Wang Haihan:
Question: prove 2 ^ (2 ^ n) - 1 | 2 ^ (2 ^ m) - 1
Let 2 ^ (2 ^ n) - 1 = x, 2 ^ (2 ^ m) - 1 = y
If x | y, then there should be x | (Y-X) and Y-X = {2 ^ (2 ^ m) - 1} - {2 ^ (2 ^ n) - 1} = 2 ^ (2 ^ m) - 2 ^ (2 ^ n) = 2 ^ (2 ^ n) {2 ^ (2 ^ m-2 ^ n) - 1} - 1
Because 2 ^ (2 ^ n) - 1 and 2 ^ (2 ^ n) are coprime, 2 ^ (2 ^ n) - 1 must be divisible by 2 ^ (2 ^ m-2 ^ n) - 1 --- 1 *
Repeat the above steps (if x|y, then x| (Y-X))
2 ^ (2 ^ n) - 1 should be divided by {2 ^ (2 ^ m-2 ^ n) - 1} - {2 ^ (2 ^ n) - 1} = 2 ^ (2 ^ n) {2 ^ (2 ^ m-2 ^ n-2 ^ n) - 1} = 2 ^ (2 ^ n) {2 ^ [2 ^ m-2 ^ n (n + 1)] - 1} -------- 2
Because 2 ^ (2 ^ n) - 1 and 2 ^ (2 ^ n) are coprime, 2 ^ (2 ^ n) - 1 must be divisible by 2 ^ [2 ^ m-2 ^ (n + 1)] - 1 ------ 2 *
By comparing Formula 1 and formula 2, it is found that the number of times of 2 changes from 2 ^ m-2 ^ n to 2 ^ m-2 ^ (n + 1). If the above steps are repeated, formula 3 can be obtained
2 ^ (2 ^ n) {2 ^ [2 ^ m-2 ^ (n + 2)] - 1} -- 3
And 2 ^ (2 ^ n) - 1 must be divisible by 2 ^ [2 ^ m-2 ^ (n + 2)] - 1 ------ 3 *
Repeat
Recursion: because n