$lg^2(10x)+lg(10x)=6-3lg(\frac{1}{x} )\\x>0\\lg^2(10x)+lg(10x)=6-3(lg(1)-lg(x))\\lg^2(10x)+lg(10x)=6-3(-lg(x))\\(lg(10)+lg(x))^2+lg(10)+lg(x)=6+3lg(x)\\1+2lg(x)+lg^2(x)+1+lg(x)=6+3lg(x)\\2+lg^2(x)=6\\lg^2(x)=4\\lg(x)=2=>x=100\\lg(x)=-2=>x=100^{-1}$

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4 года назад

Решите уравнеие ctgx-4=5tng

lesantik [10]

$\displaystyle ctgx-4=5tgx\\\\\text{OD3} :\quad \left[\begin{array}{ccc}sinx\neq0\\cosx\neq0\end{array}\right\quad \rightarrow\quad \left[\begin{array}{ccc}x\neq\pi n;\,\,n\in Z\\x\neq\frac{\pi}2+\pi n;\,\,n\in Z\end{array}\right\quad \rightarrow\quad x\neq\frac{\pi n}2;\,\,n\in Z\\\\\\\frac{1}{tgx}-4=5tgx\\\\1-4tgx=5tg^2x\\\\5tg^2x+4tgx-1=0\\\\tgx=t,\quad t\in R\\\\5t^2+4t-1=0\\\\\text{D}=16+20=36=6^2\\\\t_1=\frac{-4+6}{10}=\frac{2}{10}=\frac{1}5\\\\t_2=\frac{-4-6}{10}=-\frac{10}{10}=-1$

$\displaystyle \left[\begin{array}{ccc}\displaystyle tgx=\frac{1}5\\\\tgx=-1\end{array}\right\quad\rightarrow\quad\boxed{\left[\begin{array}{ccc}\displaystyle x=arctg\bigg(\frac{1}5\bigg)+\pi n;\,\,n\in Z\\\\\displaystyle x=-\frac{\pi}4+\pi n;\,\,n\in Z\end{array}\right}$

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4 года назад