Задание 46*

Докажите равенство:

\(\begin{array}[t]{lll}\begin{array}{l}0{,}(9)=1,\\[0.9ex]x=0{,}(9),\\[0.9ex]x=0{,}999{...},\\[0.9ex]10x=9{,}999{...},\\[0.9ex]10x-x=9,\\[0.9ex]9x=9,\\[0.9ex]x=1\end{array}&&&\begin{array}{l}{\largeа)}\ 2{,}(9)=3;\phantom{0000000000000000000000.}{\largeб)}\ 3{,}1(9)=3{,}2.&&&\\[0.9ex]\definecolor{dots}{RGB}{110,177,146}{\color{dots}{\large............................................}}\\[0.9ex]{\color{dots}{\large............................................}}\\[0.9ex]{\color{dots}{\large............................................}}\\[0.9ex]{\color{dots}{\large............................................}}\\[0.9ex]{\color{dots}{\large............................................}}\\[0.9ex]{\color{dots}{\large............................................}}\end{array}\end{array}\)

Источник заимствования: Алгебра. Рабочая тетрадь. 7 класс. В 2 частях. Часть 1 / М.К. Потапов, А.В. Шевкин – Просвещение, 2018. – 23 c. ISBN 978-5-09-051661-7

Реклама

А+АА-

Решение:

\({\largeа)}\ 2{,}(9)=3,\)
\(\phantom{\largeа)}\ x=2{,}(9),\)
\(\phantom{\largeа)}\ x=2{,}999{...},\)
\(\phantom{\largeа)}\ 10x=29{,}999{...},\)
\(\phantom{\largeа)}\ 10x-x=27,\)
\(\phantom{\largeа)}\ 9x=27,\)
\(\phantom{\largeа)}\ x=3\)
\({\largeб)}\ 3{,}1(9)=3{,}2,\)
\(\phantom{\largeб)}\ x=3{,}1(9),\)
\(\phantom{\largeб)}\ x=3{,}1999{...},\)
\(\phantom{\largeб)}\ 10x=31{,}999{...},\)
\(\phantom{\largeб)}\ 100x=319{,}999{...},\)
\(\phantom{\largeб)}\ 100x-10x=288,\)
\(\phantom{\largeб)}\ 90x=288,\)
\(\phantom{\largeб)}\ x=3{,}2\)

Задание 45*Задание 47