Решите уравнение:
\({\largeа)}\ \sqrt{\vphantom{4}3x-1}=1;\)
\({\largeб)}\ \sqrt{6x+4}=2;\)
\({\largeв)}\ \sqrt{\vphantom{4}12-x}=6;\)
\({\largeг)}\ \sqrt{\vphantom{4}8x-1}=1.\)
Упражнение 307
Источник заимствования: Математика. Алгебра. 8 класс. Базовый уровень. / Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова – Просвещение, 2023. – 74 c. ISBN 978-5-09-102536-1
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Решение:
\({\largeа)}\ \sqrt{\vphantom{4}3x-1}=1\)
\(\phantom{\largeа)}\ 3x-1=1^2\)
\(\phantom{\largeа)}\ 3x-1=1\)
\(\phantom{\largeа)}\ 3x=1+1\)
\(\phantom{\largeа)}\ 3x=2\)
\(\phantom{\largeа)}\ x=\frac{2}{3}\)
\(\phantom{\largeа)}\ 3x-1=1^2\)
\(\phantom{\largeа)}\ 3x-1=1\)
\(\phantom{\largeа)}\ 3x=1+1\)
\(\phantom{\largeа)}\ 3x=2\)
\(\phantom{\largeа)}\ x=\frac{2}{3}\)
\({\largeб)}\ \sqrt{6x+4}=2\)
\(\phantom{\largeб)}\ 6x+4=2^2\)
\(\phantom{\largeб)}\ 6x+4=4\)
\(\phantom{\largeб)}\ 6x=4-4\)
\(\phantom{\largeб)}\ 6x=0\)
\(\phantom{\largeб)}\ x=0\)
\(\phantom{\largeб)}\ 6x+4=2^2\)
\(\phantom{\largeб)}\ 6x+4=4\)
\(\phantom{\largeб)}\ 6x=4-4\)
\(\phantom{\largeб)}\ 6x=0\)
\(\phantom{\largeб)}\ x=0\)
\({\largeв)}\ \sqrt{\vphantom{4}12-x}=6\)
\(\phantom{\largeв)}\ 12-x=6^2\)
\(\phantom{\largeв)}\ 12-x=36\)
\(\phantom{\largeв)}\ x=12-36\)
\(\phantom{\largeв)}\ x={-}24\)
\(\phantom{\largeв)}\ 12-x=6^2\)
\(\phantom{\largeв)}\ 12-x=36\)
\(\phantom{\largeв)}\ x=12-36\)
\(\phantom{\largeв)}\ x={-}24\)
\({\largeг)}\ \sqrt{\vphantom{4}8x-1}=1\)
\(\phantom{\largeг)}\ 8x-1=1^2\)
\(\phantom{\largeг)}\ 8x-1=1\)
\(\phantom{\largeг)}\ 8x=1+1\)
\(\phantom{\largeг)}\ 8x=2\)
\(\phantom{\largeг)}\ x=\frac{2}{8}=\frac{1}{4}\)
\(\phantom{\largeг)}\ 8x-1=1^2\)
\(\phantom{\largeг)}\ 8x-1=1\)
\(\phantom{\largeг)}\ 8x=1+1\)
\(\phantom{\largeг)}\ 8x=2\)
\(\phantom{\largeг)}\ x=\frac{2}{8}=\frac{1}{4}\)