Help me to do some simple math problems, in which ^ 2 is the second power, ^ 3 is the third power, and 1 / 4 is a quarter ab(4a^2-5ab+b^2)-4a(a^2 b-ab^2+1/4 b^3) （2x-y)-(-2x-y) （b+2）*（2a-b） （2x^2+3y)*(2x^2-3y) (2a^2=3b)*(2a^2-3b) (x+y-1)*(x-y+1) 1998^2-1999*1997 In the triangle ABC, angle a = 108 °, angle c-angle B = 28 °, find the degree of angle B and angle C It's 4:20 this afternoon

Help me to do some simple math problems, in which ^ 2 is the second power, ^ 3 is the third power, and 1 / 4 is a quarter ab(4a^2-5ab+b^2)-4a(a^2 b-ab^2+1/4 b^3) （2x-y)-(-2x-y) （b+2）*（2a-b） （2x^2+3y)*(2x^2-3y) (2a^2=3b)*(2a^2-3b) (x+y-1)*(x-y+1) 1998^2-1999*1997 In the triangle ABC, angle a = 108 °, angle c-angle B = 28 °, find the degree of angle B and angle C It's 4:20 this afternoon

ab(4a^2-5ab+b^2)-4a(a^2 b-ab^2+1/4 b^3)
=ab(4a-b)(a-b)-4ab(2a-b/2)^2
=ab(4a-b)(a-b)-ab(4a-b)^2
=ab(4a-b)(a-b-4a+b)
=-3ba^2(4a-b)
（2x-y)-(-2x-y)
=2x-y+2x+y
=4x
（b+2）*（2a-b）
=2ab+4a^2-b^2-2b
（2x^2+3y)*(2x^2-3y)
=4x^4-9y^2
(2a^2+3b)*(2a^2-3b)
=4a^4-9b^2
(x+y-1)*(x-y+1)
=x^2-(y-1)^2
=x^2-y^2+2y-1
1998^2-1999*1997
=1998^2-(1998^2-1)
=1
In the triangle ABC, angle a = 108 °, angle c-angle B = 28 °, find the degree of angle B and angle C
Angle c + angle B = 180-108 = 72 degree
Angle c = (72 + 28) / 2 = 50 degree
Angle B = 28-50
The first is 2Ab ^ 3-5a ^ 2B ^ 2-4a ^ 2B ^ 3
The second is 4x
The third is 4AB + 4a-b ^ 2-2b
The fourth is 4x ^ 4-9y ^ 2
What does (2a ^ 2 = 3b) * (2a ^ 2-3b) mean
The fifth is x ^ 2-y ^ 2 + 2y-1
The sixth is 1
The seventh is:
In the triangle ABC, angle a = 108 °, angle c-angle B = 28 °, find the degree of angle B and angle C
Angle c + angle B = 1... Expansion
The first is 2Ab ^ 3-5a ^ 2B ^ 2-4a ^ 2B ^ 3
The second is 4x
The third is 4AB + 4a-b ^ 2-2b
The fourth is 4x ^ 4-9y ^ 2
What does (2a ^ 2 = 3b) * (2a ^ 2-3b) mean
The fifth is x ^ 2-y ^ 2 + 2y-1
The sixth is 1
The seventh is:
In the triangle ABC, angle a = 108 °, angle c-angle B = 28 °, find the degree of angle B and angle C
Angle c + angle B = 180-108 = 72 degree
Angle c = (72 + 28) / 2 = 50 degree
Angle B = 50-28 = 22 ° retract
1、-（ab）^2
2、4x
3、4a+2ab-b^2-2b
4、4x^4-9y^2
5、(2a^2-3b)*(2a^2-3b)=4a^4-12a^2*b+9b^2
6、(x+y-1)*(x-y+1)=x^2-y^2+2y-1
7、1998^2-1999*1997=1998^2-(1998+1)*(1998-1)=1998^2-1998^2+1=1
Is the 57th power of 2 minus 1 prime or composite
It must be a total!
On mathematical problems of power~~~~~~~~
15^m÷__ =5^m
3^m
15^m÷x=5^m
x=15^m÷5^m=(15÷5)^m=3^m
Is the 98th power of 2 plus 3 prime or composite
Prime number
Prime number
A mathematical problem (about solving cubic)
X³+X²+X+1=0
X³-X²-X-1=0
2X³+X²+X+1=0
X³+2X²+X+1=0
X³+X²-X+1=0
Mainly want to see the process of solving cubic
X³+X²+X+1=0
(x^3 +x^2)+(x+1)=0
x^2(x+1)+(x+1)=0
(x+1)(x^2 +1)=0
x+1=0
x=-1
x^3+x^2+x+1=0
x^2(x+1)+(x+1)=0
(x+1)(x^2 +1)=0
X ^ 2 + 1, regardless of the value of X, must be > 0
If the equation holds, only x + 1 = 0
Then x = - 1
The formula is available
X³+X²+X+1=(X³+1)+(X²+X)
=(x+1)(X²-X+1)+x(x+1)
=(x+1)(X²-X+1+x)
=(x+1)(X²+1)=0
There is only one real root: x = - 1
Let x = - 1, the above equation holds, so x + 1 is a factor of polynomial, let X & sup3; + X & sup2; + X + 1 = (x + 1) (AX & sup2; + BX + C), expand and compare the coefficients on the right side, we can see that a = 1, B = 0, C = 1. That is X & sup3; + X & sup2; + X + 1 = (x + 1) (X & sup2; + 1) = 0,
We can see x = - 1 or x = I or x = - I
Is the 66th power of 2 + 88th power of 3 prime or composite
The mantissa rule of the nth power of 2
The first power mantissa of 2 is 2
The square mantissa of 2 is 4
The third power mantissa of 2 is 8
The fourth power mantissa of 2 is 6
The fifth power mantissa of 2 is 2
The sixth power mantissa of 2 4
……
That is, the mantissa of the nth power of 2 takes 2, 4, 8 and 6 as a cycle
Again,
The mantissa rule of the nth power of 3
The first power mantissa of 3 is 3
The square mantissa of 3 is 9
The third power mantissa of 3 is 7
The fourth power mantissa of 3 is 1
The fifth power mantissa of 3 is 3
The sixth power mantissa of 3 is 9
……
That is to say, the mantissa of the nth power of 3 takes 3, 9, 7 and 1 as a cycle
According to the above rules,
The mantissa of 2 ^ 66 power is 66 △ 4 = 16 The mantissa is the second of cycles 2, 4, 8 and 6, namely 4
The mantissa of 3 ^ 88 power is 88 △ 4 = 22, and the mantissa is the fourth one in cycle 3, 9, 7 and 1, i.e. 1
In other words, if the mantissa of 2 ^ 66 + 3 ^ 88 power is 5, then the number must be divisible by 5. Besides being divisible by 1 and itself, the number can be divisible by 5 at least. According to the definition of composite number, it is a composite number
What is the 128th power of 2?
Please log it out. Thank you
Log2y 〓 128 is the power of 〓 2 〓 64 is the power of 〓 4 〓 32 is the power of 〓 16
It is known that the 859433 power - 1 of 2 is prime. Is the 859433 power + 1 of 2 prime or composite? Please explain the reason
The 859433 power of 2 is a composite number, and it must be a multiple of 2. The 859433 power of 2 - 1, the 859433 power of 2, the 859433 power of 2 + 1, these three continuous natural numbers, there must be a multiple of 3. The 859433 power of 2 - 1, is a prime number. That is to say, the 859433 power of 2 - 1 is not a multiple of 3
1、 The nth power of 2 must not be a multiple of 3
2、 2 to the power of 859433 - 1 is prime and not a multiple of 3
3、 Three continuous natural numbers must have a multiple of 3
So the 859433 power of 2 is a multiple of 3, so it's not prime
1、 The nth power of 2 must not be a multiple of 3
2、 2 to the power of 859433 - 1 is prime and not a multiple of 3
3、 Three continuous natural numbers must have a multiple of 3
So the 859433 power of 2 is a multiple of 3, so it's not prime
That's a good answer! The solution is very good in thinking. The top one!
The 859433 power of 2 + 1 is a composite number! Because the 859433 power of 2 is a composite number, and it must be a multiple of 2. The 859433 power of 2 - 1, the 859433 power of 2, the 859433 power of 2 + 1, of these three continuous natural numbers, one must be a multiple of 3. The 859433 power - 1 of 2 is prime. That is to say, the 859433 power - 1 of 2 is not a multiple of 3. The 859433 power of 2 has no factor of 3, and it must be a multiple of 2, not a multiple of 3. So the 859433 power + 1 of 2 must be the expansion of 3
The 859433 power of 2 + 1 is a composite number! Because the 859433 power of 2 is a composite number, and it must be a multiple of 2. The 859433 power of 2 - 1, the 859433 power of 2, the 859433 power of 2 + 1, of these three continuous natural numbers, one must be a multiple of 3. The 859433 power - 1 of 2 is prime. That is to say, the 859433 power - 1 of 2 is not a multiple of 3. There is no factor of 3 in the 859433 power of 2, and it must be a multiple of 2, not a multiple of 3. So the 859433 power + 1 of 2 must be a multiple of 3! It's a sum. Put it away
If m power + n of 3 can be divisible by 10, it is proved that m power + 4 power + of 3 can also be divisible by 10
It is proved that the (M + 4) power + n of 3 can also be divisible by 10
(M + 4) power of 3 = m power of 3 times 4 power of 3 = m power of 81 * 3 = m power of 3 + m power of 80 * 3
So, the (M + 4) power of < 3 + n > / 10 = the m power of < 3 + n > / 10 + the m power of 8 * 3
Among the 20 natural numbers 1-20, there are both prime and even numbers, and even odd and composite numbers
There are 0 numbers that are prime numbers and even numbers
Even if the odd number is a composite number, there are two numbers
In the natural numbers of 1-20, the odd and even prime numbers have 6, 12 and 18 multiples of 2 and 3, and 10 and 20 multiples of 2 and 5
The number that is prime and even is only 2
Even if odd numbers are combined numbers, there are 9 and 15
That is, the number that is prime and even is 2, even if the odd number is composite is 15.9
2 is prime and even, 9 and 15 are odd and composite
The number that is prime and even is only 2
Even if odd numbers are combined numbers, there are 9 and 15
1 is neither prime nor composite
The numbers that are prime numbers and even numbers are: 2
Even odd numbers are combined numbers: 9, 15