Given f (x) = 2x ^ 2 + x g (x) = 2 + 1 / x, find f (x) > G (x) g(x)=2+(1/x)

Given f (x) = 2x ^ 2 + x g (x) = 2 + 1 / x, find f (x) > G (x) g(x)=2+(1/x)

f(x)>g(x)
(2x^2+x)>2+(1/x)
x(2x+1)-(2x+1)/x>0
(2x+1)(x-1/x)>0
(2x+1)(x^2-1)/x>0
(2x+1)(x+1)(x-1)/x>0
X1 or - 1 / 2

If f (2x + 1) = 3x + 2, then f (x)=_______ ?
If f (x) defined on R satisfies f (x + y) = f (x) + F (y) + 2XY (x.y ∈ R) and f (1) = 2, then f (- 3)=_______ ?

This is one of the 2 (2 (2x + 1 / 2) 2 A: 2F (2-2 / 2) a: 2 (2-2 / 2) a: 2 (2-2 / 2) a: 2 (2-2 / 2) a: 2F (1) = f (4-3) = f (4-3) = f (4 (4-3) = f (4 (4-3-3) = 3 (3 (2x + 1 / 2) (3 (2 (2x + 1) / 2] + 1 / 2 (2 / 2 / 2) f (x) = 3x / 2 (2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2, Δ X; △ x → 0 = Lim [f (x) + F (△ x) + 2x △ x-f (x)] / △ x = 2x + limf (△ x) / △ x = 2x + F '(0) f (x) = x & sup2; + F' (0) x + C; f(0)=0,C=0 f(1)=2,f'(0)=1 f(x)=x²+x f(-3)=6

On binomial theorem
In the expansion of (x-1) (x + 1) ^ 8, the coefficient of x ^ 5 is? A specific process

First, find the coefficients of x ^ 5 and x ^ 4 in (x + 1) ^ 8, which are C85 and C84 respectively; therefore, the coefficients of x ^ 5 in the expansion are c84-c85 (note that C85 and C84 are combination numbers)

The constant term of (x + 1 / X - 2) n is - 20 (n is the nth power)
Thank you very much``

(X+1/X -2)^n=={√x-(1/√x)}^2n
C2n/n*(-1)^n=-20
{(2n)!/[n!*n!]}*(-1)^n=-20
Because 2n)! / N! * n! Is a positive number, so (- 1) ^ n is a negative number
N is odd number (2n)! / [n! * n!] = 20
The solution is n = 3

Application to binomial theorem
[1 + X + 1 / (x ^ 2)] ^ 10 finding constant term

I don't know if this is OK
In [1 + X + 1 / (x ^ 2)] ^ 10, let R + 1 be tr + 1 = C10, R × (x + 1 / x ^ 2) ^ R
In (x + 1 / x ^ 2) ^ R, let n + 1 be TN + 1 = Cr, n × x ^ (R-N) × x ^ (- 2n)
If this term is a constant term, then r-n-2n = 0 and the solution is n = R / 3
Because n is an integer, R can only be 0, 3, 6, 9
Then the constant term is c10,0 + c10,3 × c3,1 + c10,6 × c6,2 + c10,9 × c9,3 = 4351
Right? I don't know. I'll give you an idea~
P. S. can I understand the spirit of C10, 1 or something?!

In the expansion of (1 + x) 5 (1-x) 4, the coefficient of X3 is

Because the (1 + x) ^ 5 (1-x) ^ 5 ((1-x) as ((1 + x) (1-x) (1-x) (1-x (1-x) (1-x (1-x ^ 2) ^ 2 (1-x ^ 2) ^ ^ 5 ((1-x) ^ 5 * (1-x) ^ 5 (1-x) ^ 5 * (1-x) ^ 5 (1-x ((1 + x) (1 + x) (1-x) (1-x) (1-x (1-x) (1-x) (1-x) (1-x (1-x) (1-x (1-x (1-x ^ 2-x ^ 2-x ^ 2) ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 (it's not easy

In the (x + 1) ^ 4 (x-1) ^ 5 expansion, the coefficient of x ^ 4 is? And the sum of the coefficients is?

In the expansion, the coefficients of X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\8309; - 6x & # 8308; -

(x + 6) X3 = 99, X △ 7.8 = 4.1, solve the equation and test it

(x + 6) X3 = 99
3x+18=99
3x=99-18
3x=81
x=81÷3
x=27
x÷7.8=4.1
x=4.1×7.8
x=31.98

In the expansion of (x2 + 1 / x) 6, the coefficient of X3 is

Solution
(x²+1/x)^6
The coefficient of x ^ 3 is
C(6,2)(x²)^4×(1/x)²=15x^3
The coefficient is 15

The coefficient of X3 in (1 + x2) (1-x) 5 expansion is______ ．

The term containing X3 in the expansion is (- c53-c51) X3, so the coefficient of X3 is - c53-c51 = - 15, so the answer is - 15