Question

Помогите пожалуйста решить 2 и 3п пример

Answer 1

1)

$tg(-60)*ctg(300)=\frac{sin(-60)}{cos(-60)}*\frac{cos(300)}{sin(300)} \\ cos(300)=cos(240+60)=cos240cos60-sin240sin60=(-\frac{\sqrt{3}}{2}*\frac{1}{2})-(-\frac{1}{2}*\frac{\sqrt{3}}{2})=1 \\ sin(300)=sin(240+60)=sin240cos60+sin60cos240=(-\frac{\sqrt{3}}{2}*\frac{1}{2})+(\frac{\sqrt{3}}{2}*(-\frac{1}{2})=-\frac{\sqrt{3}}{2} \\ \frac{1}{\frac{-\sqrt{3}}{2}}=\frac{2}{-\sqrt{3}} \\ \frac{sin(-60)}{cos(-60)}=\frac{\sqrt{3}}{2}*(-\frac{1}{2})=-\frac{\sqrt{3}}{4} \\ (-\frac{\sqrt{3}}{4})*(-\frac{2}{\sqrt{3}})=1/2$

2)

$\frac{6}{x+1}=\frac{2}{x}+1 \\ \frac{6}{x+1}=\frac{2+x}{x} \\ 6x=(x+1)(2+x) \\ 6x=2x+x^{2}+2+x \\ x^{2}-3x+2=0 \\ x_{1}=2 \ x_{2}=1$

3)

$log_{9}3x+log_{9}1=log_{9}x \\ log_{9}3x-log_{9}x=-log_{9}1 \\ log_{9}3=-log_{9}1 \\ 3\neq0$