Найдите значение переменной \(x,\) при котором верно равенство:
\(\vphantom{\sqrt{\frac{0}{0}}}{\largeа)}\ \sqrt{3+5x}=7;\)
\(\vphantom{\sqrt{\frac{0}{0}}}{\largeб)}\ \sqrt{10x-14}=11;\)
\({\largeв)}\ \sqrt{\frac{1}{3}x-\frac{1}{2}}=0.\)
Упражнение 306
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:
\(\vphantom{\sqrt{\frac{0}{0}}}{\largeа)}\ \sqrt{3+5x}=7\)
\(\phantom{\largeа)}\ 3+5x=7^2\)
\(\phantom{\largeа)}\ 3+5x=49\)
\(\phantom{\largeа)}\ 5x=49-3\)
\(\phantom{\largeа)}\ 5x=46\)
\(\phantom{\largeа)}\ x=\frac{46}{5}\)
\(\phantom{\largeа)}\ x=9{,}2\)
\(\phantom{\largeа)}\ 3+5x=7^2\)
\(\phantom{\largeа)}\ 3+5x=49\)
\(\phantom{\largeа)}\ 5x=49-3\)
\(\phantom{\largeа)}\ 5x=46\)
\(\phantom{\largeа)}\ x=\frac{46}{5}\)
\(\phantom{\largeа)}\ x=9{,}2\)
\(\vphantom{\sqrt{\frac{0}{0}}}{\largeб)}\ \sqrt{10x-14}=11\)
\(\phantom{\largeб)}\ 10x-14=11^2\)
\(\phantom{\largeб)}\ 10x-14=121\)
\(\phantom{\largeб)}\ 10x=121+14\)
\(\phantom{\largeб)}\ 10x=135\)
\(\phantom{\largeб)}\ x=\frac{135}{10}\)
\(\phantom{\largeб)}\ x=13{,}5\)
\(\phantom{\largeб)}\ 10x-14=11^2\)
\(\phantom{\largeб)}\ 10x-14=121\)
\(\phantom{\largeб)}\ 10x=121+14\)
\(\phantom{\largeб)}\ 10x=135\)
\(\phantom{\largeб)}\ x=\frac{135}{10}\)
\(\phantom{\largeб)}\ x=13{,}5\)
\({\largeв)}\ \sqrt{\frac{1}{3}x-\frac{1}{2}}=0\)
\(\phantom{\largeв)}\ \frac{1}{3}x-\frac{1}{2}=0^2\)
\(\phantom{\largeв)}\ \frac{1}{3}x-\frac{1}{2}=0\)
\(\phantom{\largeв)}\ \frac{1}{3}x=\frac{1}{2}\)
\(\phantom{\largeв)}\ x=\frac{1}{2}:\frac{1}{3}\)
\(\phantom{\largeв)}\ x=\frac{1}{2}\cdot3\)
\(\phantom{\largeв)}\ x=\frac{3}{2}\)
\(\phantom{\largeв)}\ x=1{,}5\)
\(\phantom{\largeв)}\ \frac{1}{3}x-\frac{1}{2}=0^2\)
\(\phantom{\largeв)}\ \frac{1}{3}x-\frac{1}{2}=0\)
\(\phantom{\largeв)}\ \frac{1}{3}x=\frac{1}{2}\)
\(\phantom{\largeв)}\ x=\frac{1}{2}:\frac{1}{3}\)
\(\phantom{\largeв)}\ x=\frac{1}{2}\cdot3\)
\(\phantom{\largeв)}\ x=\frac{3}{2}\)
\(\phantom{\largeв)}\ x=1{,}5\)