Who knows the formula: x + X. (1 + 1%) + X. (1 + 1%) to the second power +. + X. (1 + 1%) to the (n-1) power Who knows the formula: x + X. (1 + 1%) + the quadratic power of X. (1 + 1%) + the (n-1) power of X. (1 + 1%) = which formula,

Who knows the formula: x + X. (1 + 1%) + X. (1 + 1%) to the second power +. + X. (1 + 1%) to the (n-1) power Who knows the formula: x + X. (1 + 1%) + the quadratic power of X. (1 + 1%) + the (n-1) power of X. (1 + 1%) = which formula,

X + X. (1 + 1%) + X. (1 + 1%) to the second power +. + X. (1 + 1%) to the (n-1) power
=x[1.01^0+1.01^1+1.01^2+.
+1.01^(n-1)]
=x[1×（1-1.01^n）]/（1-1.01）
=100x(1.01^n-1）
Equal ratio summation formula sequence Sn = [A1 (1-Q ^ n)] / (1-Q)
This is an equal ratio sequence, the first term A1 = x, the common ratio q = 1 + 1%
Substitution formula
S=(a1)*(1-q^n)/(1-q)
=x*[1-(1+1%)^n]/[1-(1+1%)]
=[x-x*1.01^n]/(-1%)
=100x*(1.01^n-1)
[(a + b) (a-b) - (a-b) square + 2B (a-b)] divided by 4 (a-b)
The process should be complete
The original formula = (A & sup2; - B & sup2; - A & sup2; + 2ab-b & sup2; + 2ab-2b & sup2;) △ 4 (a-b)
=(-4b²+4ab)÷4(a-b)
=4b(a-b)÷4(a-b)
=b
Factorization and integral division of eight
To make a judgment (or correct a mistake)
①a5+a5=a10 ( ) ②(x3)5=x8( ) ③a3×a3= a6 ( )
④y7y=y8( ) ⑤a3×a5= a15 ( ) ⑥(x2)3 x4 = x9( )
⑦b4×b4= 2b4 ( ) ⑧(xy3)2=xy6( ) ⑨（－2x）5 = －2x3( )
(- 2XY) 4 = (3a2) n = (x4) 6 - (x3) 8 = (TM) 2 times t
① A 5 + a 5 = a 10 (x) 2 (x 3) 5 = x 8 (x) 3 a 3 × a 3 = a 6 (pair) 4 y 7 y = y 8 (pair) 5 a 3 × a 5 = a 15 (x) 6 (x 2) 3 x 4 = x 9 (x) 7 B 4 × B 4 = 2 B 4 (x) 8 (x y 3) 2 = x y 6 (x) 9 (- 2x) 5 = - 2x 3 (x) (- 2XY) 4 = 4 of the fourth power y of 16 X
(1) Wrong. The correct answer is: A5 + A5 = 2A.
(2) Wrong. The answer to ^ 3 is (x = 5).
(3) Yes.
(4) Yes.
(5) Wrong. The correct answer is: A ^ 3xa ^ 5 = a ^ 8.
(6) Wrong. The correct answer is: (x ^ 2) ^ 3x ^ 4 = x ^ 10.
(7) Wrong. The correct answer is: B ^ 4xb ^ 4 = B ^ 8.
The solution of the equation to the x power of 3 + the Y power of 4 = the Z power of 5 is x = y = z = 2
Some other Pythagorean numbers also satisfy the solution. What's the relationship between them? Please prove it!
Please give strict proof! Why does the equation have a unique solution 2
If there are other solutions, it is better to assume that there is the same increment t on the basis of 2, t is not equal to 0, then there is: 3 ^ (2 + T) + 4 ^ (2 + T) = 5 ^ (2 + T) 9 * 3 ^ t + 16 * 4 ^ t = 25 * 5 ^ t; if this formula is true, there must be: 3 ^ t = 4 ^ t = 5 ^ t; and regard it as a function
Because Pythagorean theorem is 3 square + 4 square = 5 square, the solution of the equation is x = y = z = 2, that is, 3 square + 4 square = 5 square is the same
If this is the problem of divide by one, is it beyond the scope
The square of a-2ab = 4, the square of b-ab = 4 find the square of 2b-a
b² - ab = 4
Multiply both sides by 2 at the same time
2b² - 2ab = 8
Because a & # 178; - 2Ab = 4
The result of subtracting the two formulas is as follows:
2b² - a² = 4
It's four
The square of B - AB = 4 is multiplied by 2; minus the square of a - 2Ab = 4; it is exactly equal to the square of 2B - the square of a; that is, it is equal to 4 * 2-4 = 4;
Give it to me. Thank you for your cooperation!
If we know the square of B - AB = 4 and multiply both sides by 2, we can get
The square of 2b-2ab = 8, that is, the square of 2b-8 = 2Ab
From the first known equation: the square of a - 4 = 2Ab
So: the square of 2B - 8 = the square of a - 4,
The result is: the square of 2B - the square of a = 8-4 = 4
Factorization
(m²+n²)²-4m²n² x ²+2x+1-y² (a+1)²(2a-3)-2(a+1)(3-2a)+2a-3 x²-2xy+y²-2x+2y+1
(m²+n²)²-4m²n² =m²n²(m²n²-4)=[(mn)²-2²]=m²n²(mn+2)(mn-2)x ²+2x+1-y² =(x+1)²-y²=（x+1+y)(x+1-y)(a+1)²(2a-3)-2(a...
(m²-n²)²
（x-y+1)(x+y+1)
(a+2)²(2a-3)
(x-y-1)²
(m²+n²)²-4m²n² =(m^2-n^2)=(m+n) ²(m-n)²
x ²+2x+1-y²=(x+1) ²-y ²=(x+1-y)(x+1+y)
(a + 1) & #178; (2a-3) - 2 (a + 1) (3-2a) + 2a-3 = (2a-3) [(a + 1) & #178; + 2... Expand
(m²+n²)²-4m²n² =(m^2-n^2)=(m+n) ²(m-n)²
x ²+2x+1-y²=(x+1) ²-y ²=(x+1-y)(x+1+y)
(a+1)²(2a-3)-2(a+1)(3-2a)+2a-3 =(2a-3)[(a+1)²+2(a+1)+1]=(2a-1)(a+2)²
x²-2xy+y²-2x+2y+1=(x-y)²-2(x-y)+1=(x-y-1)²
Mathematics is not easy to answer, do their own, please accept satisfaction Oh, put away
(m²+n²)²-4m²n²
=（m²+n²+2mn）（m²+n²-2mn）
=（m+n）²（m-n）²
x ²+2x+1-y²
=（x+1）²-y²
=（x+y+1）（x-y+1）
2X cubic + 7 = 0 is a cubic binomial equation with one variable, and its solution is
2X cubic + 7 = 0
x³=-7/2
x=-³√(7/2)
x=-³√(28/8)
x=-(³√28)/2
x³=-7/2
So another real root
2X ^ 3 = - 7. So x = (- 7) ^ (1 / 3)
Square of (a + 2b) - square of 4 (a-b)
All the formulas for multiplication, division and factorization of integers in Volume 1 of Grade 8 are listed one by one,
1：（a＋b）²＝a²＋2ab＋b²2: （a－b）²＝a²－2ab＋b²3: a²－b²＝（a－b）（a＋b）4: （a＋b＋c）²＝a²＋b²＋c²＋2ab＋2ac＋2bc5: （a＋b）³...
If we know that the | a | + 1 power of equation (A-2) x = 0 is a linear equation of one variable with respect to x, then a = ()
The | a | + 1 power of (A-2) x = 0 is a linear equation of one variable with respect to X
The power exponent | a | + 1 = 1, so | a | = 0, a = 0
a-2≠0|a|+1=1a≠2a=0
If the | a | + 1 power of equation (A-2) x = 0 is a linear equation of one variable with respect to x, then a = (0)