(1,5a⁻²b)(4a³b⁻⁴) = 6a⁻²⁺³b¹⁻⁴ = 6ab⁻³ = 6a/b³;
Тримай розв'язання. Якщо щось буде незрозумілим — запитуй, відповім.
![\sqrt{x+3}- \sqrt{3x-2}\ \textgreater \ \sqrt{x-2} \\](https://tex.z-dn.net/?f=+%5Csqrt%7Bx%2B3%7D-+%5Csqrt%7B3x-2%7D%5C+%5Ctextgreater+%5C++%5Csqrt%7Bx-2%7D+%5C%5C++++)
ОДЗ:
![x+3 \geq 0 \\ 3x-2 \geq 0 \\ x-2 \geq 0 \\ \\ x \geq 2](https://tex.z-dn.net/?f=x%2B3+%5Cgeq+0+%5C%5C+3x-2+%5Cgeq+0+%5C%5C+x-2+%5Cgeq+0+%5C%5C++%5C%5C+x+%5Cgeq+2)
Возведем в квадрат
![x+3-2 \sqrt{(x+3)(3x-2)} -3x+2\ \textgreater \ x-2 \\ 2 \sqrt{(x+3)(3x-2)}\ \textless \ -3x+7](https://tex.z-dn.net/?f=x%2B3-2+%5Csqrt%7B%28x%2B3%29%283x-2%29%7D+-3x%2B2%5C+%5Ctextgreater+%5C+x-2+%5C%5C+2+%5Csqrt%7B%28x%2B3%29%283x-2%29%7D%5C+%5Ctextless+%5C+-3x%2B7)
Повторно возведем в квдрат
![4(x+3)(3x-2)\ \textless \ (-3x+7)^2 \\ 12x^2-32x+16\ \textless \ 9x^2-42x+49 \\ 3x^2+10x-33\ \textless \ 0 \\ \\ 3x^2+10x-33=0 \\ D=100+396=496=(4 \sqrt{31})^2 \\ x_{1,2}= \dfrac{-10б4 \sqrt{31} }{6} = \dfrac{-5б2 \sqrt{31} }{3}](https://tex.z-dn.net/?f=4%28x%2B3%29%283x-2%29%5C+%5Ctextless+%5C+%28-3x%2B7%29%5E2+%5C%5C+12x%5E2-32x%2B16%5C+%5Ctextless+%5C+9x%5E2-42x%2B49+%5C%5C+3x%5E2%2B10x-33%5C+%5Ctextless+%5C+0+%5C%5C++%5C%5C+3x%5E2%2B10x-33%3D0+%5C%5C+D%3D100%2B396%3D496%3D%284+%5Csqrt%7B31%7D%29%5E2+%5C%5C+x_%7B1%2C2%7D%3D+%5Cdfrac%7B-10%D0%B14+%5Csqrt%7B31%7D+%7D%7B6%7D++%3D+%5Cdfrac%7B-5%D0%B12+%5Csqrt%7B31%7D+%7D%7B3%7D+)
a>0 ⇒ x∈(-5-2√31/3;-5+2√31/3)
С учетом ОДЗ
x∈[2;-5+2√31/3)
Ответ: x∈[2;-5+2√31/3)
По теореме косинусов
![BC^2=AB^2+AC^2-2AB\cdot AC\cdot \cos \angle A\\ \\ BC^2=4^2+6^2-2\cdot 4\cdot 6\cdot \cos 60^\circ\\ \\ BC^2=16+36-2\cdot 24\cdot \dfrac{1}{2}\\ \\ BC^2=28\\ \\ BC=2\sqrt{7}](https://tex.z-dn.net/?f=BC%5E2%3DAB%5E2%2BAC%5E2-2AB%5Ccdot%20AC%5Ccdot%20%5Ccos%20%5Cangle%20A%5C%5C%20%5C%5C%20BC%5E2%3D4%5E2%2B6%5E2-2%5Ccdot%204%5Ccdot%206%5Ccdot%20%5Ccos%2060%5E%5Ccirc%5C%5C%20%5C%5C%20BC%5E2%3D16%2B36-2%5Ccdot%2024%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%5C%5C%20%5C%5C%20BC%5E2%3D28%5C%5C%20%5C%5C%20BC%3D2%5Csqrt%7B7%7D)
По теореме синусов
![\dfrac{BC}{\sin \angle A}=2R~~~\Rightarrow~~~ R=\dfrac{BC}{2\sin 60^\circ}=\dfrac{2\sqrt{7}}{\sqrt{3}}=\boxed{\dfrac{2\sqrt{21}}{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7BBC%7D%7B%5Csin%20%5Cangle%20A%7D%3D2R~~~%5CRightarrow~~~%20R%3D%5Cdfrac%7BBC%7D%7B2%5Csin%2060%5E%5Ccirc%7D%3D%5Cdfrac%7B2%5Csqrt%7B7%7D%7D%7B%5Csqrt%7B3%7D%7D%3D%5Cboxed%7B%5Cdfrac%7B2%5Csqrt%7B21%7D%7D%7B3%7D%7D)
Площадь треугольника ABC:
кв. ед.
С другой стороны
отсюда и выразим радиус вписанной окружности
![r=\dfrac{2S}{P}=\dfrac{2\cdot 6\sqrt{3}}{4+6+2\sqrt{7}}=\boxed{\dfrac{6\sqrt{3}}{5+\sqrt{7}}}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7B2S%7D%7BP%7D%3D%5Cdfrac%7B2%5Ccdot%206%5Csqrt%7B3%7D%7D%7B4%2B6%2B2%5Csqrt%7B7%7D%7D%3D%5Cboxed%7B%5Cdfrac%7B6%5Csqrt%7B3%7D%7D%7B5%2B%5Csqrt%7B7%7D%7D%7D)