Question

3^x+2 + 9^x+1 - 810 = 0



lim   x/корен3+x - корень3-x
x->0

Answer 1

$3^{x+2} + 9^{x+1} - 810 = 0\3*3^{x+1}+3^{2(x+1)}-810=0\3^{2y}+3*3^y-810=0; (y=x+1)\(3^y+30)(3^y-27)=0\3^y+30>0 forall y\3^y = 27=3^3\y=3\x=y-1=2$

$lim_{x to 0}(frac{x}{sqrt{3+x}-sqrt{3-x}})=lim_{x to 0}(frac{x(sqrt{3+x}+sqrt{3-x})}{(sqrt{3+x}-sqrt{3-x})(sqrt{3+x}+sqrt{3-x})})=\=lim_{x to 0}frac{x(sqrt{3+x}+sqrt{3-x})}{2x}=frac{2sqrt{3}}{2}=sqrt{3}$