How to compare the size of X X 1 and X 1 X, X 3, and X is an integer

Answer: Conclusion: x (x+1) (x+1) x

Proof is as follows: x (x+1) (x+1) x

(x+1)lnx＞xln(x+1)

(lnx)/x＞[ln(x+1)]/(x+1)..

Therefore, it is only necessary to consider the monotonicity of the function f(x)=lnx x.

f(x)=[(1 x)*x-1nx] x 2=(1-lnx) x 2, obviously, when x 3, f(x) 0, f(x) is a subtractive function on (3,+).

Establish. i.e. when x 3 there is: x (x+1) (x+1) x

Consider the function f(x)=(1+1 x) x

f'(x)=[ln(1+1 x)-1 (1+x)](1+1 x) x, then let g(x)=ln(1+1 x)-1 (1+x)g'(x)=-x (1+x) 2<0(x>0), i.e. g(x) is a subtraction function at (0, + is the subtraction function.

and x, g(x) 0

Thus g(x) is at (0,+ on Evergrande to 0

i.e. f'(x) at (0, + on Evergrande at 0

i.e. f(x) is an increasing function on (0,+).

and x, f(x) e

Thus f(x) is (1+1 x) x x and both sides are multiplied by x x.

x+1) xps: Remember to use the elementary method to prove.

f(x)<3, which is also sufficient to prove (x+1) x

The method is: ln(x+1) x-lnx (x+1)=xln(x+1)-(x+1)lnx

xln(x+1)-xlnx-lnx=xln((x+1)/x)-lnx

When x is large, the former term tends to be close to 0, and the difference between the two is negative, so when (x+1) x=3, x(x+1)>(x+1) x

Compare 3 4 and 4 3 and calculate the values to compare.

If you compare x (x+1) and (x+1) x is a bit more cumbersome.

Solution: (x+1)(x+4)-(3x +4x+5)=x +5x+4-3x -4x-5=-2x Split+x-1=-(2x -x+1)=-2(x-1 4) +7 8] 0, (x+1)(x+4) 3x Leak+4x+5

Summary. The result of the difference between the two polynomials is 2,2 0, so the former and the latter.

Compare the size of x-x with (x+1) (x-2).

Look at the **, dear. Compare the size of the two numbers, make the difference directly, and look at the relationship between the difference and 0.

The result of the difference between the two polynomials is 2,2 0, so the former and the latter.

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Subtracting the two polynomials yields: (x +x+3)-(x +x-1)=4 From this, we can see that x +x+3 is greater than x +x-1 and the difference is 4. Since both polynomials are of 3 degrees and their coefficients are positive, we only need to compare the magnitude of their coefficients, i.e., the coefficients of x, the coefficients of x, and the coefficients of constant terms.

When the coefficients of x and x are the same, the magnitude of the two polynomials can be determined by comparing the magnitude of the coefficients of the constant term. In this problem, the coefficients of the constant terms are -1 and 3, respectively, and 3 is greater than -1, so x +x+3 is greater than x +x-1.

Compare the size of x+x-1 with x+x+3.

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Subtracting the two multinomial configurations to obtain: (x +x+3)-(x +x-1)=4 From this, it can be seen that x +x+3 is greater than x +x-1, and the difference is 4. Since the magnitude of both polynomials is 3 times, and the coefficients of the two polynomials are positive, we only need to compare the magnitude of their coefficients, i.e., the coefficients of x, the coefficients of x, and the coefficients of the constant terms.

When the coefficients of x and x are the same, the magnitude of the coefficients of the constant term can be used to determine the size of the two multinomial land socks. In this problem, the coefficients of the constant terms are -1 and 3, respectively, and 3 is greater than -1, so x +x+3 is greater than x +x-1.

You can write it out in a notebook.

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Solving and analytic solutions of polynomials: For some complex polynomial equations, it is necessary to find analytical solutions or approximate solutions, which is of great significance for the research and practical application of Komo nucleus. 3.

Polynomial interpolation and approximation: Using the known polynomial function values, a curve or surface is constructed, so that the curve or surface is as close as possible to the known function values, so as to obtain more accurate and accurate function values. 4.

Zui optimization problem of polynomials: For some polynomial functions, it is necessary to solve the large or small value of zui under certain conditions, which involves the research and application of optimization problems.

x- (x-1) =1 ( x+ (x-1)) multiply x+ (x-1) and divide).

x+1)- x = 1 ( x+1)+x) (Same as above).

The denominator on the left is small, so the value on the left side of the book is large.

Because the bend of the group is x 1, x square 1

x-square-1 0, x-square-1 less than x-squared.

x-squared-1) is less than x, i.e. (x+1) (x-1) is less than x(x+1)+ collapsed (x-1)] squared = 2x+2 (x+1) (x-1) is less than 4x

x+1)+x-1)] square less than 2 x squared.

x+1) + x-1) is less than 2 x

i.e. split (x+1)- x is less than x- (x-1).