[tex]1.\quad \Big(\sqrt{5\cdot3}\Big)^2=5\cdot3=15[/tex]
[tex]2. \quad \sqrt{2\dfrac1{12}\cdot48}=\sqrt{\dfrac{2\cdot12+1}{12}\cdot48}=\sqrt{25\cdot4}=5\cdot2=10[/tex]
[tex]3.\quad \dfrac{\sqrt{108}}{\sqrt{12}}=\sqrt{\dfrac{108}{12}}=\sqrt9=3[/tex]
[tex]4.\quad \Big(\sqrt{98}-5\sqrt{2}\Big)\sqrt2=\Big(7\sqrt2-5\sqrt2\Big)\sqrt2=7\cdot2-5\cdot2=4[/tex]
Ответ: 1 - В; 2 - А; 3 - Г; 4 - Д (ВАГД)
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
[tex]1.\quad \Big(\sqrt{5\cdot3}\Big)^2=5\cdot3=15[/tex]
[tex]2. \quad \sqrt{2\dfrac1{12}\cdot48}=\sqrt{\dfrac{2\cdot12+1}{12}\cdot48}=\sqrt{25\cdot4}=5\cdot2=10[/tex]
[tex]3.\quad \dfrac{\sqrt{108}}{\sqrt{12}}=\sqrt{\dfrac{108}{12}}=\sqrt9=3[/tex]
[tex]4.\quad \Big(\sqrt{98}-5\sqrt{2}\Big)\sqrt2=\Big(7\sqrt2-5\sqrt2\Big)\sqrt2=7\cdot2-5\cdot2=4[/tex]
Ответ: 1 - В; 2 - А; 3 - Г; 4 - Д (ВАГД)