Given that 3x + 2Y = 44, 2x + 3Y = 46, then x=______ ，y=______ ．

Given that 3x + 2Y = 44, 2x + 3Y = 46, then x=______ ，y=______ ．

The two sides of the equation are multiplied by 2, 6x + 4Y = 88 (1; 2x + 3Y = 46; the two sides of the equation are multiplied by 3, 6x + 9y = 138, 6x + 9y = 138, and the two sides of the equation are multiplied by 3, 6x + 6x + 9y = 138, and the two sides of the equation are multiplied by 2, 6x + 4Y = 88; the two sides of the equation are multiplied by 2, 6x + 6x + 4Y = 88 (1; 2x2x + 3x + 3x + 3Y = 88Y = 88, and the two sides of the equation are multiplied by 3, 6x + 6x + 6x + 9x + 9y = 3-9y = 3, which is the two of the equation of the two-2x + 2x + 6x + 6x + 6x + 6x + 6y = 88, which is the equation, and the equation of the equation of the equation of the equation of the equation of the equation of the equation of the equation of the equation& nbsp; & nbsp; & nbsp; X = 8, so the answer is: 8, 10

How to solve 2Y + 3x = 46, 3Y + 2x = 44?

The result is 6y + 9x = 138
The second formula * 2 gives 6y + 4x = 88
By subtracting the two formulas, we can get 5x = 50 and x = 10
Y = 8 is obtained by substituting
That's it. It's done

Given that 3x + 2Y = 44, 2x + 3Y = 46, then x=______ ，y=______ ．

3x + 2Y = 44, two sides of the equation multiply by 2, 6x + 4Y = 88, ①; 2x + 3Y = 46, two sides of the equation multiply by 3, 6x + 9y = 138, ②; ② subtract ①, 5Y = 138-885y = 50 & nbsp; y = 10; substitute y into 3x + 2Y = 44, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 3x + 20 = 44 & nbsp; & nbsp; 3x + 20 = 44-2

The seventh power of (x-1) is equal to the second power of Ao + a1x + a2x +... + the seventh power of a7x?
Speed (Mathematics of grade one)

When x = - 1
a0+a1x+a2x^2+.a7x^7 =a0-a1+a2-a3+...-a7=-(2)^7
When x = 1
a0+a1x+a2x^2+.a7x^7 =a0+a1+a2+a3+.a7=0
Add the two formulas to get A0 + A2 + A4 + A6 = 2 ^ 6

Let (3-x) be n-th power = A0 + a1x + a2x & # 178; + +Anx, and A0 + A1 + A2 + +An = 32, then A3 =?

Let x = 1
The n-th power of (3-1) = A0 + A1 + A2 + +an
N power of 2 = 32
n=5
The fourth term = C (5,3) * 3 & # 178; * (- x) &# 179; = - 90x & # 179;
a3=-90
Please click here if you are satisfied
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Given (2x-1) 5 = a5x5 + a4x4 + a3x3 + a2x2 + a1x + A0, then the value of A2 + A4 is ()
A. -120B. 140C. -140D. 70

Let x = 0, get - 1 = A0; let x = 1, get 1 = A5 + A4 + a3 + A2 + A1 + A0 ①; let x = - 1, get - A5 + a4-a3 + a2-a1 + A0 = - 243 ②; let 1 = 1, get - A5 + a4-a3 + a2-a1 + A0 = - 243 ②; let 1 + ②, get 2a4 + 2A2 + 2a0 = - 242, that is, A0 + A2 + A4 = - 121, - 1 + A2 + A4 = - 121, - A2 + A4 = - 120

The power field is expanded to x = a

a^x=e^u ,u=x*ln a
Expand e ^ u according to e ^ x formula, and then substitute u into it
The convergence domain is infinite

In the summation of power series, how to operate term by term derivation or term by term integration?
For example: ∑ (n from 1 to positive infinity) (- 1) ^ (n-1) * (2x) ^ (2n-1) / (2n-1)
Σ (n from 1 to positive infinity) n * (n + 2) x ＾ 2n

① Σ (n from 1 to positive infinity) (- 1) ^ (n-1) * (2x) ^ (2n-1) / (2n-1)
=Σ (n from 1 to positive infinity) (- 1) ^ (n-1) ∫ (2x) ^ (2n-2) DX (integral interval is 0 to x, the same below)
=Σ (n from 1 to positive infinity) (- 1) ^ (n-1) ∫ (4x & # 178;) ^ (n-1) DX
=Σ (n from 1 to positive infinity) ∫ (- 4x & # 178;) ^ (n-1) DX
=∫ [∑ (n from 1 to positive infinity) (- 4x & # 178;) ^ (n-1)] DX
=∫[1/(1+4x²)]dx
=arctan2x
② Σ (n from 1 to positive infinity) n * (n + 2) x ＾ 2n
=1 / 2 ∑ (n from 1 to positive infinity) 2n (n + 2) x ^ 2n
=(1 / 2) x ∑ (n from 1 to positive infinity) (n + 2) 2nx ^ (2n-1)
=(1 / 2) x ∑ (n from 1 to positive infinity) (n + 2) [x ^ (2n)] '
=(1 / 2) x [∑ (n from 1 to positive infinity) (n + 2) x ^ (2n)] '
Σ (n from 1 to positive infinity) (n + 2) x ^ (2n)
=1 / (2x & # 179;) ∑ (n from 1 to positive infinity) (2n + 4) x ^ (2n + 3)
=1 / (2x & # 179;) ∑ (n from 1 to positive infinity) [x ^ (2n + 4)] '
=1 / (2x & # 179;) [∑ (n from 1 to positive infinity) x ^ (2n + 4)] '
=1/(2x³)[x^6/(1-x²)]′
=x²(3-2x²)/(1-x²)²
The original formula = (1 / 2) x [∑ (n from 1 to positive infinity) (n + 2) x ^ (2n)] '
=(1/2)x[x²(3-2x²)/(1-x²)²]′
=x²(3-x²) /(1-x)³

1. Expand the function f (x) = x ^ 6 + 2x ^ 4-x + 1 into a polynomial of X-1 according to Taylor series. 2. Expand a ^ x into a power series of X

1) Let t = X-1, then x = t + 1, and substitute f (x) f (x) = (T + 1) ^ 6 + 2 (T + 1) ^ 4 - (T + 1) + 1 = (T + 1) ^ 6 + 2 (T + 1) ^ 4-T expansion to get the polynomial about t, which is the expansion about X-1. This is a polynomial, finite term. 2) e ^ x = 1 + X + x ^ 2 / 2! + x ^ 3 / 3! +... A ^ x = e ^ (xlna) = 1 + xlna + (xlna) ^ 2 / 2! + (x

Who can give me some examples of Taylor series and power series expansion, and talk about the difference between them
Can I write it in PDF? No, just read it

Examples are really hard to write. I'll show you their relationship in words
There are many concepts here. We need to understand the power series, the power series expansion of Taylor series function of convergence function of power series
1. Power series is a large range, Taylor series is relative to a function f (x)
You can write a series of x ^ n, no matter how you write the coefficient, it is a power series
The Taylor series is the product of the N derivative of F (x) at x0 and (x-x0) ^ n, and then divided by n! N from 0 to infinity. This series is called the Taylor series of F (x) at x0. Note here that... Is the Taylor series of F (x) at x0, So it's also a power series!
Power series expansion and power series are two different things!
The number of power series is a series, but the statement of power series expansion is not accurate. It should be said that the expansion of power series of F (x) is to expand f (x) into the form of power series, find a power series, what kind of power series? If the sum function of the power series converges to f (x), how to find it? Check the Taylor series of F (x) in x0, when does the series converge to f (x), If the Lagrangian remainder of F (x) tends to infinity, the Taylor series of F (x) at x0 converges to f (x), so f (x) develops into his Taylor series. This expansion is unique
So the power series expansion of F (x) is to make the Taylor series of F (x) converge at a certain point, and then the power series expansion of F (x) is found
Played a lot, I don't know if you can understand, I hope you can read it carefully
In fact, there are similar concepts. Fourier series of function and Fourier expansion of function are also different concepts
PS: the one on the first floor is bullshit. The series of function in x = 0 is called McLaughlin series of function. It's a Taylor series of function! The concept is not clear!