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Реши уравнение

Creato il: 15 agosto 2025

Risposto usando GPT-5 da Chat01

Domanda

Screenshot_20250815-225428.pngScreenshot_20250815-225428.png

Реши уравнение

Python

python
from PIL import Image, ImageOps, ImageFilter, ImageEnhance, ImageDraw, ImageFont, ImageStat, Image
img=Image.open('/mnt/data/Screenshot_20250815-225428.png')
img.size

(1080, 2400)

Python

python
display(img.crop((130,830,1020,1110)).resize((900,300)))

Python

python
display(img.crop((140,1080,1020,1280)).resize((900,250)))

Python

python
import mpmath as mp

f=lambda x: mp.atan((2*x-1)/(x+2)) + mp.atan((x+3)/(3*x-1)) - x
# avoid x where denom zero: x!=-2 and x!=1/3
for guess in [ -3,-1,0,0.1,0.5,1,2,3,4]:
    try:
        r= mp.findroot(f, guess)
        print(guess, r)
    except: 
        pass

Pensiero

Investigating potential roots

The roots I found are approximately -1.7127 and 1.4289. However, I need to check for extraneous solutions, especially due to the arctan branch behavior. Since the sum of arctans is confined within the range (-π, π), these roots seem valid. To solve analytically, we might try using the tangent of both sides.

Applying tangent to both sides

I need to apply the tangent function to both sides. The identity for tan(arctan A + arctan B) is (A+B)/(1 - AB). The equation will become tan(x) = [(2x-1)/(x+2) + (x+3)/(3x-1)] / [1 - ((2x-1)/(x+2)) * ((x+3)/(3x-1))]. I'll simplify it, making sure x stays in the range (-π, π) for equality. It'll still be a transcendental equation though.

Thought for 54s

Risposta

Нужно решить

arctan⁡2x−1x+2+arctan⁡x+33x−1=x.\arctan\frac{2x-1}{x+2}+\arctan\frac{x+3}{3x-1}=x .

Сложим арктангенсы через формулу:

tan⁡ ⁣(arctan⁡a+arctan⁡b)=a+b1−ab.\tan\!\left(\arctan a+\arctan b\right)=\frac{a+b}{1-ab}.

Тогда

tan⁡x=2x−1x+2+x+33x−11−2x−1x+2⋅x+33x−1=7.\tan x=\frac{\frac{2x-1}{x+2}+\frac{x+3}{3x-1}} {1-\frac{2x-1}{x+2}\cdot\frac{x+3}{3x-1}} =7.

Значит

x=arctan⁡7+kπ.x=\arctan 7+k\pi.

Но левая часть — сумма главных значений арктангенсов — всегда лежит в интервале (− π,π)(-\,\pi,\pi). Поэтому возможны только те xx, которые попадают в этот интервал:

 x=arctan⁡7илиx=arctan⁡7−π .\boxed{\,x=\arctan 7\quad \text{или}\quad x=\arctan 7-\pi\, }.

Численно:

x≈1.42889927илиx≈−1.71269338.x\approx 1.42889927 \quad \text{или}\quad x\approx -1.71269338.

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