Question

Решите номер 5 .Есть вложение. 25 б

Answer 1

$(frac{1}{sqrt3}ln(sqrt3x+sqrt{3x^2-2})'=frac{1}{sqrt3}cdot(ln(sqrt3x+sqrt{3x^2-2})'=$

$frac{1}{sqrt3}cdotfrac{1}{sqrt3x+sqrt{3x^2-2}}cdot(sqrt3x+sqrt{3x^2-2})'=$

$frac{1}{sqrt3}cdotfrac{1}{sqrt3x+sqrt{3x^2-2}}cdotleft(sqrt3+ frac{1}{2sqrt{3x^2-2}} cdot(3x^2-2)'right) =$

$frac{1}{sqrt3cdot(sqrt3x+sqrt{3x^2-2})}cdotleft(sqrt3+ frac{6x}{2sqrt{3x^2-2}}right) =$

$frac{1}{sqrt3cdot(sqrt3x+sqrt{3x^2-2})}cdotfrac{sqrt3cdot2sqrt{3x^2-2}+6x}{2sqrt{3x^2-2}} =$

$frac{1}{sqrt3cdot(sqrt3x+sqrt{3x^2-2})}cdotfrac{2sqrt3(sqrt{3x^2-2}+sqrt3x)}{2sqrt{3x^2-2}} =$

$frac{1}{sqrt{3x^2-2}}$

Answer 2

mioment error

Answer 3

уже ok

Answer 4

Спасибо большое, https://znanija.com/task/31725559 помогите с алгеброй 25 б