Реши

Создано: 10 февраля 2026 г.

Отвечено с использованием GPT-5.2 Thinking от Chat01

Вопрос

IMG_0335.jpeg

Реши

Python

python
from PIL import Image, ImageEnhance
img = Image.open('/mnt/data/IMG_0335.jpeg')
img.size


(4032, 3024)

Python

python
import matplotlib.pyplot as plt, numpy as np
rot=img.rotate(90, expand=True)
rot.size


(3024, 4032)

Python

python
# Let's crop region with writing left side?
import numpy as np
arr=np.array(rot)
# show with matplotlib in analysis (not visible to user but helpful)
plt.figure(figsize=(6,8))
plt.imshow(rot)
plt.axis('off')
plt.show()

Python

python
rot2=rot.rotate(180, expand=True)
plt.figure(figsize=(10,4))
plt.imshow(rot2)
plt.axis('off')
plt.show()

Python

python
# crop top band
w,h=rot2.size
crop=rot2.crop((0,0,w,int(h*0.25)))  # top quarter
plt.figure(figsize=(12,3))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
# crop region including expression (approx y=200..1200)
crop=rot2.crop((0,150,rot2.size[0],1200))
plt.figure(figsize=(12,4))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
crop=rot2.crop((0,150,rot2.size[0],1700))
plt.figure(figsize=(14,5))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
import matplotlib.pyplot as plt
from PIL import ImageEnhance, ImageOps
# further crop around full fraction
frac=rot2.crop((0,300,rot2.size[0],1500))
# enhance contrast
gray=ImageOps.grayscale(frac)
enh=ImageEnhance.Contrast(gray).enhance(2.5)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(14,4))
plt.imshow(enh, cmap='gray')
plt.axis('off')
plt.show()

Python

python
right=rot2.crop((1900,420,3024,1050))
gray=ImageOps.grayscale(right)
enh=ImageEnhance.Contrast(gray).enhance(3.0)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(8,4))
plt.imshow(enh, cmap='gray')
plt.axis('off')
plt.show()

Python

python
right=rot2.crop((1600,650,3024,1250))
gray=ImageOps.grayscale(right)
enh=ImageEnhance.Contrast(gray).enhance(3.0)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(10,4))
plt.imshow(enh, cmap='gray')
plt.axis('off')
plt.show()

Python

python
arg=rot2.crop((1220,520,2050,1200))
gray=ImageOps.grayscale(arg)
enh=ImageEnhance.Contrast(gray).enhance(3.0)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(10,4))
plt.imshow(enh, cmap='gray')
plt.axis('off')
plt.show()

Python

python
left=rot2.crop((0,430,900,1050))
gray=ImageOps.grayscale(left)
enh=ImageEnhance.Contrast(gray).enhance(3.0)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(10,4))
plt.imshow(enh, cmap='gray')
plt.axis('off')
plt.show()

Python

python
den=rot2.crop((900,1100,1500,1600))
gray=ImageOps.grayscale(den)
enh=ImageEnhance.Contrast(gray).enhance(3.0)
enh=ImageEnhance.Sharpness(enh).enhance(2.0)
plt.figure(figsize=(8,3))
plt.imshow(enh, cmap='gray')
plt.axis('off')
plt.show()

Python

python
import sympy as sp
sqrt=sp.sqrt
x = sqrt(2)/(3*sqrt(2)-1)
B = 25**(sp.log(x, sp.Rational(1,5))) * 2**(sp.log(5**sp.Rational(1,3), 5))
A = sp.log(3 + 1/sqrt(2), 3)
C = 3**(sp.log(10, 5))  # log base 5 of 10
expr = (A - B) / C
sp.nsimplify(expr), sp.N(expr,20)


(3**(-1 - log(2)/log(5))*(-2**(log(5**(1/3))/log(5))/25**(log(sqrt(2)/(-1 + 3*sqrt(2)))/log(5)) + log(sqrt(2)/2 + 3)/log(3)),
 -1.1279525026881477853)

Python

python
B_simpl = (19-6*sqrt(2))/2 * 2**sp.Rational(1,3)
sp.simplify(B - B_simpl), sp.N(B_simpl,20)


(0, 6.6238576651592592368)

Python

python
sp.N(A,20), sp.N(C,20)


(1.1926425205254761750, 4.8151097955721152523)

Мысль

Re-evaluating the expression and terms

Okay, so I got an answer of -1.12795, but that doesn't feel right. I'm starting to think I might have misread the denominator: Perhaps it's not an exponent, but multiplication, like $3 \times \log_5(10)$ . Yet, the handwriting suggests an exponent. I should stick with $3^{\log_5(10)}$ . Let’s keep checking.

Simplifying terms and checking expressions

Okay, so B simplifies to $x^{-2} \cdot 2^{1/5}$ , still irrational. But maybe $25^{\log_{1/5} (\sqrt{2}/(3\sqrt{2}-1))}$ simplifies to $x^{-2}$ — that’s rational with a square root. Still, logs and radicals complicate the numerator. Let's keep the denominator $3^{\log_5 10}$ , for now and check the final simplified expression.

Simplifying expressions and matching terms

It looks like 25 equals $(1/5)^{-2}$ , and then with the simplifications, the expression becomes $x^{-2} = ((3\sqrt{2}-1)/\sqrt{2})^2 = (19-6\sqrt{2})/2$ . The numerator involves a logarithm term, and the denominator could potentially simplify, though it seems complex. Let’s proceed with final simplification and check the details.

Thought for 4m 53s

Ответ

С картинки читается выражение

\frac{\log_3\!\left(3+\frac1{\sqrt2}\right)\;-\;25^{\log_{1/5}\left(\frac{\sqrt2}{3\sqrt2-1}\right)}\cdot 2^{\log_5(\sqrt[3]{5})}}{3^{\log_5 10}}.

Упростим множители в числителе.

1) $25^{\log_{1/5}x}$

Пусть $x=\dfrac{\sqrt2}{3\sqrt2-1}$ . Заметим, что $25=\left(\frac15\right)^{-2}$ . Тогда

25^{\log_{1/5}x}=\left(\left(\frac15\right)^{-2}\right)^{\log_{1/5}x} =\left(\frac15\right)^{-2\log_{1/5}x} =\Big(\left(\tfrac15\right)^{\log_{1/5}x}\Big)^{-2} =x^{-2}.

Значит,

25^{\log_{1/5}\left(\frac{\sqrt2}{3\sqrt2-1}\right)} =\left(\frac{\sqrt2}{3\sqrt2-1}\right)^{-2} =\left(\frac{3\sqrt2-1}{\sqrt2}\right)^2 =\frac{(3\sqrt2-1)^2}{2} =\frac{19-6\sqrt2}{2}.

2) $2^{\log_5(\sqrt[3]{5})}$

\log_5(\sqrt[3]{5})=\log_5(5^{1/3})=\frac13 \quad\Rightarrow\quad 2^{\log_5(\sqrt[3]{5})}=2^{1/3}=\sqrt[3]{2}.

Итог

Выражение равно

\boxed{ \frac{\log_3\!\left(3+\frac1{\sqrt2}\right)\;-\;\frac{19-6\sqrt2}{2}\,\sqrt[3]{2}} {3^{\log_5 10}} }

Численно это примерно:

\boxed{-1.1279525027}.

Если хочешь — посчитаю точнее или проверю, не иначе ли читается знаменатель (там похоже именно $3^{\log_5 10}$ ).

Вопрос

В числителе 2^log домножается не только на второй но и на первый многочлен

Python

python
from PIL import Image, ImageEnhance
img = Image.open('/mnt/data/IMG_0335.jpeg')
img.size


(4032, 3024)

Python

python
import matplotlib.pyplot as plt, numpy as np, PIL
# rotate to make expression horizontal
rot = img.rotate(90, expand=True)
rot.size


(3024, 4032)

Python

python
# crop region containing expression (left side after rotation)
import numpy as np
arr = np.array(rot)
# Let's inspect by cropping a vertical strip where writing is
crop = rot.crop((0,0,900,4032))  # left 900 px
crop.size


(900, 4032)

Python

python
# enhance contrast and convert
gray = crop.convert('L')
enh = ImageEnhance.Contrast(gray).enhance(2.5)
enh = ImageEnhance.Sharpness(enh).enhance(2.0)
# further crop around middle where formula is
mid = enh.crop((0,500,900,1900))
mid.size


(900, 1400)

Python

python
import matplotlib.pyplot as plt
plt.figure(figsize=(10,6))
plt.imshow(mid, cmap='gray')
plt.axis('off')
plt.show()

Python

python
# Let's show whole left strip scaled down to locate formula
small = enh.resize((300, int(enh.size[1]*300/enh.size[0])))
plt.figure(figsize=(4,10))
plt.imshow(small, cmap='gray')
plt.axis('off')
plt.show()

Python

python
bottom = enh.crop((0,2500,900,4032))
small2 = bottom.resize((900, int(bottom.size[1]*900/bottom.size[0])))
plt.figure(figsize=(12,6))
plt.imshow(small2, cmap='gray')
plt.axis('off')
plt.show()

Python

python
rot2 = img.rotate(-90, expand=True)
rot2.size


(3024, 4032)

Python

python
strip = rot2.crop((0,0,900,4032)).convert('L')
strip_enh = ImageEnhance.Contrast(strip).enhance(2.5)
strip_enh = ImageEnhance.Sharpness(strip_enh).enhance(2.0)
small = strip_enh.resize((300, int(strip_enh.size[1]*300/strip_enh.size[0])))
plt.figure(figsize=(4,10))
plt.imshow(small, cmap='gray')
plt.axis('off')
plt.show()

Python

python
top = strip_enh.crop((0,0,1800,1200))
plt.figure(figsize=(12,4))
plt.imshow(top, cmap='gray')
plt.axis('off')
plt.show()

Python

python
# create downscaled full rotated image for inspection
full = rot2.convert('L')
full_enh = ImageEnhance.Contrast(full).enhance(2.0)
full_enh = ImageEnhance.Sharpness(full_enh).enhance(2.0)
small_full = full_enh.resize((800, int(full_enh.size[1]*800/full_enh.size[0])))
plt.figure(figsize=(8,10))
plt.imshow(small_full, cmap='gray')
plt.axis('off')
plt.show()

Python

python
# crop around numerator to inspect
crop2 = full_enh.crop((0,650,3000,1200))
plt.figure(figsize=(14,3))
plt.imshow(crop2, cmap='gray')
plt.axis('off')
plt.show()

Python

python
den = full_enh.crop((500,1050,1800,1500))
plt.figure(figsize=(10,3))
plt.imshow(den, cmap='gray')
plt.axis('off')
plt.show()

Python

python
import sympy as sp
sqrt2=sp.sqrt(2)
x = sqrt2/(3*sqrt2-1)
expr_25 = 25**(sp.log(x, sp.Rational(1,5))) # sympy log(a,b) means log base b? Actually sympy log(x, b) is log(x)/log(b).
expr_25_simplified = sp.simplify(expr_25)
expr_25_simplified


289/(sqrt(2) + 6)**2

Python

python
sp.simplify(expr_25_simplified - (19-6*sqrt2)/2)


0

Python

python
A = sp.log(3+1/sqrt2, 3)  # log base 3
B = (19-6*sqrt2)/2
C = 2**sp.Rational(1,3)
D = 3**(sp.log(10,5))
expr = (A - B) * C / D
sp.N(expr, 20)


-1.0635733069386940045

Мысль

A = sp.log(3+1/sqrt2, 3) # log base 3 B = (19-6*sqrt2)/2 C = 2sp.Rational(1,3) D = 3(sp.log(10,5)) expr = (A - B) * C / D sp.N(expr, 20)

-1.0635733069386940045

Thought for 3m 43s

Ответ

Понял: в числителе стоит скобка, и $2^{\log_5(\sqrt[3]{5})}$ умножает всю разность:

E=\frac{\Big(\log_3\!\Big(3+\frac1{\sqrt2}\Big)-25^{\log_{1/5}\left(\frac{\sqrt2}{3\sqrt2-1}\right)}\Big)\cdot 2^{\log_5(\sqrt[3]{5})}}{3^{\log_5 10}}.

1) Упростим $25^{\log_{1/5}x}$

Пусть $x=\dfrac{\sqrt2}{3\sqrt2-1}$ . Заметим $25=\left(\frac15\right)^{-2}$ . Тогда

25^{\log_{1/5}x}=\left(\left(\frac15\right)^{-2}\right)^{\log_{1/5}x} =\left(\frac15\right)^{-2\log_{1/5}x} =\Big(\left(\tfrac15\right)^{\log_{1/5}x}\Big)^{-2}=x^{-2}.

Значит

25^{\log_{1/5}\left(\frac{\sqrt2}{3\sqrt2-1}\right)} =\left(\frac{\sqrt2}{3\sqrt2-1}\right)^{-2} =\left(\frac{3\sqrt2-1}{\sqrt2}\right)^2 =\frac{(3\sqrt2-1)^2}{2} =\frac{19-6\sqrt2}{2}.

2) Упростим $2^{\log_5(\sqrt[3]{5})}$

\log_5(\sqrt[3]{5})=\log_5(5^{1/3})=\frac13 \quad\Rightarrow\quad 2^{\log_5(\sqrt[3]{5})}=2^{1/3}=\sqrt[3]{2}.

Итог

\boxed{ E=\frac{\sqrt[3]{2}}{3^{\log_5 10}} \left(\log_3\!\left(3+\frac1{\sqrt2}\right)-\frac{19-6\sqrt2}{2}\right) }

Численно:

E\approx \boxed{-1.0635733069}.

Вопрос

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Мысль

Re-evaluating the expression and terms

Simplifying terms and checking expressions

Simplifying expressions and matching terms

Ответ

1) 25log⁡1/5x25^{\log_{1/5}x}25log1/5​x

2) 2log⁡5(53)2^{\log_5(\sqrt[3]{5})}2log5​(35​)

Итог

Вопрос

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Мысль

Ответ

1) Упростим 25log⁡1/5x25^{\log_{1/5}x}25log1/5​x

2) Упростим 2log⁡5(53)2^{\log_5(\sqrt[3]{5})}2log5​(35​)

Итог

Поделиться этим Q&A

1) $25^{\log_{1/5}x}$

2) $2^{\log_5(\sqrt[3]{5})}$

1) Упростим $25^{\log_{1/5}x}$

2) Упростим $2^{\log_5(\sqrt[3]{5})}$