F (x) = (x + 1) / the limit of (x ^ 2 + 1) when x tends to infinity

F (x) = (x + 1) / the limit of (x ^ 2 + 1) when x tends to infinity

One
Factorization of 4x2 - 9y2 - (2x + 3Y)
Factorization question 1: 4x2 - 9y2 - (2x + 3Y)
Question 2: x3-x2-xy-y2-y3
Question 3: x3-2x2 + 2x-1
Question 4: A2-B2 + B-1 / 4
Question 5: x2 + 2XY + Y2 + 7x + 7y-18
Question 1: 4x2 - 9y2 - (2x + 3Y)
=(2x+3y)(2x-3y)-(2x+3y)
=(2x+3y)(2x-3y-1)
Question 2: x3-x2-xy-y2-y3
=x³-y³-(x²+xy+y²)
=(x²+xy+y²)(x-y-1)
Question 3: x3-2x2 + 2x-1
=x³-1+2x(x-1)
=(x-1)(x²+3x+1)
Question 4: A2-B2 + B-1 / 4
=a²-(b²-b+1/4)
=(a+b-1/2)(a-b+1/2)
Question 5: x2 + 2XY + Y2 + 7x + 7y-18
=(x+y)²+7(x+y)-18
=(x+y+9)(x+y-2)
When x tends to 1, both f (x) = X-1 / x + 1 and G (x) = root X-1 are infinitesimal. This paper compares the infinitesimal order of F (x) and G (x)
lim(x→1)f(x)/g(x)=lim(x→1)√(x-1)/(x+1)=0
So f (x) is infinitesimal of higher order than g (x)
7：x=14：（2+x）
7(2+x)=14x
14+7x=14x
14x-7x=14
7x=14
x=14÷7
X=2
The original formula = (X-2) &# 178; / (X-2) - (X & # 178; - 5x) / (x-3)
=x-2-(x²-5x)/(x-3)
=[(x-2)(x-3)-(x²-5x)]/(x-3)
=(x²-5x+6-x²+5x)/(x-3)
=6/(x-3)
Factorization 121 - (x + 2) ²
121-（x+2）²
＝11²-（x+2）²
=(11+x+2)(11-x-2)
=(13+x)(9-x)
It is known that f (x) = a (x-1) ^ 2 + B (x-1) + C-radical (x ^ 2 + 3) is the infinitesimal of higher order of (x-1) ^ 2 when x approaches 1
Find the value of a, B and C. I just missed it
I may have miscalculated, but the idea is right. Just expand by Taylor under the root (x ^ 2 + 3)
[decompose the following formulas into factors, 25Q & # 178; - 121p & # 179;
Let's factorize the following,
0.25q²-121p³
(49 / 4) a-178; - x-178; y-178;
9x²-（2y+z）²
49（2a-3b）²-9（a+b）²
We all use the square difference formula: 0.25q \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\i'd like to
0.25q²-121p²
=(0.5q+11p)(0.5q-11p)
(49 / 4) a & # 178; - X & # 178; Y & # 178; = (7a / 2 + XY) (7a / 2-xy)
9x²-（2y+z）²=(3x+2y+z)(3x-2y-z)
49（2a-3b）²-9（a+b）²
=[7(2a-3b)+3(a+b)][7(2a-3b)-3(a+b)]
=(17a-18b)(11aa-24b)
Solving indefinite integrals with the method of partial integration: ∫ [(1 + SiNx) / (1 + cosx)] * e ^ x * DX
∫e^x*(1+sinx)/(1+cosx)dx
=∫e^x/(1+cosx)dx+∫e^xsinx/(1+cosx)dx
=∫e^x/(1+cosx)d+∫sinx/(1+cosx)de^x
=∫e^x/(1+cosx)d+e^xtan(x/2)-∫e^x/(1+cosx)dx （sinx/(1+cosx)=tan(x/2)）
=(e^x)tan(x/2) + C
121 (a-b) & sup2; - 169 (a + b) & sup2;, factorization
It is helpful for the responder to give an accurate answer
Because 121 = 11 * 11169 = 13 * 13, the square of 121 (a-b) (a-b) - 169 (a + b) (a + b) = 11 * (a-b) - 13 * (a + b) is determined by the square difference formula a * A-B * b = (a + B) (a-b) original formula = [11 * (a-b) + 13 * (a + b)] [11 (a-b) - 13 (a + b)] = (24a + 2b) (- 2a-24b) = - 2 (12a + b) (a + 12b)
121（a-b）^2-169（a+b）^2
=[12(a-b)+13(a+b)][12(a-b)-13(a+b)]
=(12a-12b+13a+13b)(12a-12b-13a-13b)
=(25a+b)(-a-25b)
Indefinite integral ∫ 1 / (SiNx ^ 2 + 2cosx ^ 2) DX
∫ 1 / (sin²x + 2 cos²x) dx
=∫ (1 / cos²x) / (2 + tan²x) dx
Let u = TaNx, Du = sec & # 178; X DX
=∫ 1 / (2 + u²) du
Let u = √ 2 tanv, Du = √ 2 sec & # 178; V DV
2 + u² = 2 + 2 tan²v = 2 sec²v
=(√2 / 2) ∫ 1 / sec²v * sec²v dv
=1 / √2 * v+C
=(1 / √2) arctan(u / √2) + C
=(1 / √2) arctan[tanx / √2] + C
∫ 1 / (sin²x + 2cos²x ) dx
= ∫ (1 / sin²x) / (1 + 2cot²x ) dx
= - ∫ 1 / (1 + 2cot²x ) dcotx
= - 1 / √2 ∫ 1 / [1 + (√2cotx )²] d√2cotx
= - 1 / √2 arctan(√2 cotx) + C
You can divide cosx ^ 2 by the numerator and denominator, take the upper 1 / cosx ^ 2 as dtanx, and the lower 2 + TaNx ^ 2. The rest is very simple...
Factorization a ^ 2 + 2Ab + B ^ 2-121
The second problem, known that a = x / 2, B is a polynomial, in the calculation of B + A, a student regarded B + A as B △ a, the result is 4x & # 178; + X, can you find B + a?
a^2+2ab+b^2-121
=(a+b)²-121
=(a+b+11)(a+b-11)
A=x/2
B÷A=2B/x=4x²+x
B=2x³+x²/2
B+A=2x³+x²/2+x/2
(a+b+11)^2(a+b-11)^2
　　a^2+2ab+b^2-121
=(a+b)^2-11^2
　　=(a+b+11)(a+b-11)
　　(2)B=(4x²+x)×x/2=2x^3+(x^2)/2
　　B+A=2x^3+(x^2)/2+x/2