$1)\; \; \lim\limits _{x \to 0} \frac{sin(5\sqrt[3]{x})\cdot ln(1+\sqrt[6]{64x})}{arctg\sqrt{25x}}=\lim\limits _{x \to 0} \frac{5\sqrt[3]{x}\cdot 2\sqrt[6]{x}}{5\sqrt{x}}= \lim\limits _{x \to 0}\; (2x^{ \frac{1}{3}+\frac{1}{6}-\frac{1}{2}})=\\\\= \lim\limits _{x \to 0} (2x^0)=2\\\\2)\; \; \lim\limits _{x \to 2}\frac{3-\sqrt{x^2+5}}{x^3-8}= \lim\limits _{x \to 2}\frac{(3-\sqrt{x^2+5})(3+\sqrt{x^2+5})}{(3+\sqrt{x^2+5})(x-2)(x^2+2x+4)} =\\\\= \lim\limits _{x \to 2}\frac{9-(x^2+5)}{(3+ \sqrt{x^2+5})(x-2)(x^2+2x+4)}=\lim\limits _{x \to 2}\, \frac{4-x^2}{(3+ \sqrt{x^2+5})(x-2)(x^2+2x+4)}=$

$= \lim\limits _{x \to 2}\, \frac{(2-x)(2+x)}{-(2-x)(x^2+2x+4)(3+ \sqrt{x^2+5} )}= \lim\limits _{x \to 2}\frac{2+x}{-(x^2+2x+4)(3+ \sqrt{x^2+5} )} =\\\\= \frac{4}{-12\cdot 6} =- \frac{1}{18}$

0 0

4 года назад

6/(x+1)(x+2)+8/(x-1)(x+4)=1

Алена Мальцева [166]

(x+1)*(x+2) = x^2 +3x+2;
(x-1)*(x+4) = x^2+3x - 4;
x^2 + 3x = y;
(6/(y+2)) + (8/(y-4)) = 1;
y не=-2 и y не=4;
6*(y-4) + 8*(y+2) = (y+2)*(y-4);
6y-24 + 8y +16 = y^2 - 2y -8;
y^2 -y*(2+8+6) - 8 + 24 -16 = 0;
y^2 - 16y =0;
y*(y-16) = 0;
y=0 или y-16=0;
1) y=0; <=> x^2+3x = 0; <=> x*(x+3) = 0; <=> x1=0; или x2=-3.
2) y-16 = 0; <=> x^2+3x - 16 = 0;
D = 3^2 + 4*16 = 9+64 = 73;
x3 = (-3+(V73))/2;
x4 = (-3-(V73))/2.

0 0

4 года назад