Given that f (x) = LNX, G (x) = 13x3 + 12x2 + MX + N, the image of line L and function f (x), G (x) are tangent to point (1,0) (1) find the equation of line L and the analytic expression of G (x); (2) if h (x) = f (x) - G '(x) (where G' (x) is the derivative of G (x), find the range of function H (x)

Given that f (x) = LNX, G (x) = 13x3 + 12x2 + MX + N, the image of line L and function f (x), G (x) are tangent to point (1,0) (1) find the equation of line L and the analytic expression of G (x); (2) if h (x) = f (x) - G '(x) (where G' (x) is the derivative of G (x), find the range of function H (x)

(1) The line L is the tangent of the function f (x) = LNX at the point (1,0), so its slope k = f '(1) = 1, so the equation of the line L is y = X-1. (2 points) and because the line L is tangent to the image of G (x), the derivative value of G (x) = 13x3 + 12x2 + MX + n at the point (1,0) is 1. G (1) = 0g' (1) = 1 {M = − 1n = 16, so g (x) = 13x3 + 12x2 − x + 16 (6 points) (2) because H (x) = F (x) - G ′ (x) = lnx-x2-x + 1 (x > 0) (7 points) so h ′ (x) = 1x − 2x − 1 = 1 − 2x2 − XX = − (2x − 1) (x + 1) x (9 points) when 0 < x < 12, H ′ (x) > 0; when x > 12, H ′ (x) < 0 (11 points) therefore, when x = 12, H (x) gets the maximum value, H (12) = 14 − LN2 (12 points) so the range of function H (x) is (− ∞, 14 − LN2]. (13 points)

What is the minimum positive period of the function f (x) = SiN x times cos x
2 / 2 or 4

Original formula = 1 / 2sin2x
So period T = Wu

Reasoning 1 / 5 - () 1 / 5 = () () () / ()
Headache ah, primary school problems, and can not use the University method

1 / 5 - (0) 1 / 5 = (1) 1 / (5)
1 / 5 - (1) 1 / 5 = (0) 1 / (5)

Y = 1 / 3x & # 179; + X & # 178; + X's pole x = - 1 is a false proposition, please explain the reason

y'=x²+2x+1=(x+1)²
Then X-1 has y '> - 0
So both sides are increasing
So x = - 1 is not an extreme point

How to prove that y = x & # 179; + ax & # 178; + BX + C is a centrosymmetric graph

Its center of symmetry is on the function, and its abscissa is - B / 3a. It's easier for you to understand if you have learned derivative and integral
If proved, f (x) = x & # 179; + ax & # 178; + BX + C
Let two points (- B / 3A + T, f (- B / 3A + T)), (- B / 3a-t, f (- B / 3a-t))
f(-b/3a+t)-f(-b/3a)
=at^3+[3a*b^2/9a^2+2b*(-b/3a)+c]t
Similarly,
f(-b/3a-t)-f(-b/3a)
=-at^3-[3a*b^2/9a^2+2b*(-b/3a)+c]t
So f (- B / 3A + T) - f (- B / 3a) = f (- B / 3a-t) - f (- B / 3a)
So (- B / 3a, f (- B / 3a)) is the center of symmetry
I have a question

Calculate 1 / 1 × 3 + 1 / 3 × 5 + 1 / 5 × 7 + +1/1997×1999

1/1×3+1/3×5+1/5×7+… +1/1997×1999
=1/2[1/1-1/3+1/3-1/5+1/5-1/7+...+1/1997-1/1999]
=1/2[1-1/1999]
=999/1999

Simple calculation of 199.8 times 6 plus 9.99 times 80

199.8×6+9.99×80
=（200-0.2）×6+（10-0.01）×80
=200×6-0.2×6+10×80-0.01×80
=120-1.2+800-0.8
=120+800-1.2-0.8
=920-2
=918

How much is 1880 multiplied by 199.9 minus 1999 multiplied by 187.9
What I want is not a direct calculation, but a calculation process

1880×199.9－1999×187.9
=10×188×199.9-10×199.9×187.9
=10×199.9×（188－187.9）
=10×199.9×0.1
=199.9

(873*477-198)/(476*874+199))

Hello
(873*477-198)/(476*874+199)
=[(874-1)*477-198)/[(477-1)*874+199]
=(874*477-477-198)/(477*874-874+199)
=(874*477-675)/(874*477-675)
=1
I don't understand. I can ask
If you have any help, please remember to adopt it. Thank you
I wish you progress in your study!

(476 * 873 + 675) / (476 * 874 + 199) simple calculation

Original formula = (476 * 873 + 675) / (476 * 873 + 476 + 199)
=(476*873+675)/(476*873+675)
=1