Найдите значение переменной \(x\), при котором:
\({\largeа)}\ \sqrt{x}=4;\)
\({\largeб)}\ \sqrt{x}=0{,}5;\)
\({\largeв)}\ 2\sqrt{x}=0;\)
\({\largeг)}\ 4\sqrt{x}=1;\)
\({\largeд)}\ \sqrt{x}-8=0;\)
\({\largeе)}\ 3\sqrt{x}-2=0.\)
Упражнение 303
Источник заимствования: Математика. Алгебра. 8 класс. Базовый уровень. / Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова – Просвещение, 2023. – 73 c. ISBN 978-5-09-102536-1
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Решение:
\({\largeа)}\ \sqrt{x}=4\)
\(\phantom{\largeа)}\ x=4^2\)
\(\phantom{\largeа)}\ x=16\)
\(\phantom{\largeа)}\ x=4^2\)
\(\phantom{\largeа)}\ x=16\)
\({\largeб)}\ \sqrt{x}=0{,}5\)
\(\phantom{\largeб)}\ x=0{,}5^2\)
\(\phantom{\largeб)}\ x=0{,}25\)
\(\phantom{\largeб)}\ x=0{,}5^2\)
\(\phantom{\largeб)}\ x=0{,}25\)
\({\largeв)}\ 2\sqrt{x}=0\)
\(\phantom{\largeв)}\ \sqrt{x}=0\)
\(\phantom{\largeв)}\ x=0^2\)
\(\phantom{\largeв)}\ x=0\)
\(\phantom{\largeв)}\ \sqrt{x}=0\)
\(\phantom{\largeв)}\ x=0^2\)
\(\phantom{\largeв)}\ x=0\)
\({\largeг)}\ 4\sqrt{x}=1\)
\(\phantom{\largeг)}\ \sqrt{x}=\frac{1}{4}\)
\(\phantom{\largeг)}\ x=\left(\frac{1}{4}\right)^2\)
\(\phantom{\largeг)}\ x=\frac{1}{16}\)
\(\phantom{\largeг)}\ \sqrt{x}=\frac{1}{4}\)
\(\phantom{\largeг)}\ x=\left(\frac{1}{4}\right)^2\)
\(\phantom{\largeг)}\ x=\frac{1}{16}\)
\({\largeд)}\ \sqrt{x}-8=0\)
\(\phantom{\largeд)}\ \sqrt{x}=8\)
\(\phantom{\largeд)}\ x=8^2\)
\(\phantom{\largeд)}\ x=64\)
\(\phantom{\largeд)}\ \sqrt{x}=8\)
\(\phantom{\largeд)}\ x=8^2\)
\(\phantom{\largeд)}\ x=64\)
\({\largeе)}\ 3\sqrt{x}-2=0\)
\(\phantom{\largeе)}\ 3\sqrt{x}=2\)
\(\phantom{\largeе)}\ \sqrt{x}=\frac{2}{3}\)
\(\phantom{\largeе)}\ x=\left(\frac{2}{3}\right)^2\)
\(\phantom{\largeе)}\ x=\frac{4}{9}\)
\(\phantom{\largeе)}\ 3\sqrt{x}=2\)
\(\phantom{\largeе)}\ \sqrt{x}=\frac{2}{3}\)
\(\phantom{\largeе)}\ x=\left(\frac{2}{3}\right)^2\)
\(\phantom{\largeе)}\ x=\frac{4}{9}\)