Question

Решить уравнение
$\frac{(39-x)\sqrt[5]{x-6}-(x-6)\sqrt[5]{39-x}}{\sqrt[5]{39-x}-\sqrt[5]{x-6}}=30$

Answer 1

Ответ: 7; 38

Объяснение:

$Let \ \sqrt[5]{39-x}=a, \ \sqrt[5]{x-6}=b, \ x\ne22.5\\\dfrac{a^5b-ab^5}{a-b}=30\\\dfrac{ab(a^4-b^4)}{a-b}=30\\\dfrac{ab(a^2+b^2)(a+b)(a-b)}{a-b}=30\\ab((a+b)^2-2ab)(a+b)=30\\Let \ a+b=c, \ ab=d\\cd(c^2-2d)=30\\c^3d-2cd^2=30\\11c^3d-22cd^2=330\\Zametim .. ))\\a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)=(a+b)(a^4+2a^2b^2+b^4-a^3b-a^2b^2-ab^3)=(a+b)((a^2+b^2)^2-ab(a^2+ab+b^2))=(a+b)(((a+b)^2-2ab)^2-ab((a+b)^2-ab))=c((c^2-2d)^2-d(c^2-d))=c(c^4-4c^2d+4d^2-c^2d+d^2)=c(c^4-5c^2d+5d^2)=c^5-5c^3d+5cd^2=33$

$10c^5-50c^3d+50cd^2=330\\10c^5-50c^3d+50cd^2=11c^3d-22cd^2\\10c^5-61c^3d+72cd^2=0\\c(10c^4-61c^2d+72d^2)=0\\Let \ \dfrac{c^2}{d}=t\\c(10t^2-61t+72)=0\\D=61^2-4\cdot10\cdot72=3721-2880=841=29^2\\t_1=\dfrac{61+29}{20}=\dfrac92\\t_2=\dfrac{61-29}{20}=\dfrac85\\1) c=0\\a+b=0\\a=-b\\\sqrt[5]{39-x}=\sqrt[5]{6-x}\\\varnothing$

$2) 2c^2=9d\\2(a+b)^2=9ab\\2a^2+4ab+2b^2=9ab\\2a^2-5ab+2b^2=0\\D=25-16=9\\\dfrac{a}{b}=\dfrac{5+3}{4}=2\\a=2b\\\sqrt[5]{39-x}=2\sqrt[5]{x-6}\\39-x=32x-192\\33x=231\\x=7\\\dfrac{a}{b}=\dfrac{5-3}{4}=\dfrac12\\2a=b\\2\sqrt[5]{39-x}=\sqrt[5]{x-6}\\1248-32x=x-6\\33x=1254\\x=38$

$3) 5c^2=8d\\5(a+b)^2=8ab\\5a^2+10ab+5b^2=8ab\\5a^2+2ab+5b^2=0\\D<0\\\varnothing$