Question

Высшая математика решите плиз

Answer 1

$y= \frac{\sqrt{2x+3}(x-2)}{x^2} \\\\y'= \frac{(\frac{2}{2\sqrt{2x+3}}(x-2)+\sqrt{2x+3})x^2-\sqrt{2x+3}(x-2)\cdot 2x}{x^4} =\\\\= \frac{ \frac{x(x-2)}{\sqrt{2x+3}}+x\sqrt{2x+3}-2\cdot (x-2)\sqrt{2x+3}}{x^3}= \frac{x^2-2x+x(2x+3)-(2x-4)(2x+3)}{x^3\sqrt{2x+3}}=\\\\= \frac{-x^2+3x+12}{x^3\sqrt{2x+3}}$