$\sqrt[3]{x-9}+2x^2+3x=$$\sqrt[]{5x-1}+1$

Đáp án:

`\root{3}{x - 9} + 2x^2 + 3x =\sqrt{5x -1} +1 ( ĐK: x≥ 1/5)`$\\$ `<=> \root{3}{x - 9} + 2x^2 + 3x- \sqrt{5x -1} - 1= 0` $\\$ `<=> 2\root{3}{x - 9} + 4x^2 + 6x -2\sqrt{5x -1} -2 = 0`$\\$ `<=> 2 ( \root{3}{x -9}+2) + ( 5x -1 - 2\sqrt{5x -1}) + 4x^2 + x -5 = 0`$\\$ `<=> (2(x-1))/((\root{3}{x-9})^2 - 2\root{3}{x-9}+ 4) + (5(x-1)\sqrt{5x-1})/(\sqrt{5x -1} +2) + ( x -1) ( 4x +5) = 0`$\\$`<=> (x -1)( 2/((\root{3}{x -9} -1)^2 +3) +(5\sqrt{ 5x -1})/(\sqrt{5x -1} +2) + 4x + 5) = 0`$\\$ `<=> x = 1`