Câu 3: Trang 21 sách phát triển năng lực toán lớp 9 tập 1Tính:a)$\sqrt{50}:\sqrt{18}$;...

{
"content1": {
"a": "a) $\\sqrt{50} : \\sqrt{18} = \\sqrt{\\frac{50}{18}} = \\sqrt{\\frac{25}{9}} = \\frac{5}{3}$",
"b": "b) $\\sqrt{\\frac{7}{6}} : \\sqrt{\\frac{14}{9}} = \\sqrt{\\frac{7}{6} : \\frac{14}{9}} = \\sqrt{\\frac{7}{6} \\times \\frac{9}{14}} = \\sqrt{\\frac{3}{4}} = \\frac{\\sqrt{3}}{2}$",
"c": "c) $(7\\sqrt{32} + 3\\sqrt{18} - 2\\sqrt{8}) : \\sqrt{2} = \\frac{7\\sqrt{32}}{\\sqrt{2}} + \\frac{3\\sqrt{18}}{\\sqrt{2}} - \\frac{2\\sqrt{8}}{\\sqrt{2}} = 7\\sqrt{16} + 3\\sqrt{9} - 2\\sqrt{4} = 7\\times 4 + 3\\times 3 - 2\\times 2 = 28 + 9 - 4 = 33$",
"d": "d) $\\frac{\\sqrt{18}}{\\sqrt{2}} - \\frac{\\sqrt{12}}{\\sqrt{3}} = \\frac{\\sqrt{18} \\times \\sqrt{3}}{\\sqrt{2} \\times \\sqrt{3}} - \\frac{\\sqrt{12} \\times \\sqrt{2}}{\\sqrt{3} \\times \\sqrt{2}} = \\frac{\\sqrt{54}}{\\sqrt{6}} - \\frac{\\sqrt{24}}{\\sqrt{6}} = \\frac{3\\sqrt{6}}{\\sqrt{6}} - \\frac{2\\sqrt{6}}{\\sqrt{6}} = 3 - 2 = 1$"
},
"content2": {
"a": "a) $\\sqrt{50} : \\sqrt{18} = \\sqrt{\\frac{50}{18}} = \\sqrt{\\frac{25 \\times 2}{9}} = \\frac{5\\sqrt{2}}{3}$",
"b": "b) $\\sqrt{\\frac{7}{6}} : \\sqrt{\\frac{14}{9}} = \\sqrt{\\frac{7}{6} : \\frac{14}{9}} = \\sqrt{\\frac{7 \\times 9}{6 \\times 14}} = \\sqrt{\\frac{63}{84}} = \\frac{3\\sqrt{7}}{2\\sqrt{3}}$",
"c": "c) $(7\\sqrt{32} + 3\\sqrt{18} - 2\\sqrt{8}) : \\sqrt{2} = 7\\sqrt{16}\\sqrt{2} + 3\\sqrt{9}\\sqrt{2} - 2\\sqrt{4}\\sqrt{2} = 7\\times 4\\sqrt{2} + 3\\times 3\\sqrt{2} - 2\\times 2\\sqrt{2} = 28\\sqrt{2} + 9\\sqrt{2} - 4\\sqrt{2} = 33\\sqrt{2}$",
"d": "d) $\\frac{\\sqrt{18}}{\\sqrt{2}} - \\frac{\\sqrt{12}}{\\sqrt{3}} = \\sqrt{\\frac{18}{2}} - \\sqrt{\\frac{12}{3}} = \\sqrt{9} - \\sqrt{4} = 3 - 2 = 1$"
},
"content3": {
"a": "a) $\\sqrt{50} : \\sqrt{18} = \\frac{\\sqrt{50}}{\\sqrt{18}} = \\frac{\\sqrt{25} \\times \\sqrt{2}}{\\sqrt{9}} = \\frac{5\\sqrt{2}}{3}$",
"b": "b) $\\sqrt{\\frac{7}{6}} : \\sqrt{\\frac{14}{9}} = \\frac{\\sqrt{7} \\div \\sqrt{6}}{\\sqrt{14} \\div \\sqrt{9}} = \\frac{\\sqrt{7} \\times \\sqrt{9}}{\\sqrt{6} \\times \\sqrt{14}} = \\frac{3\\sqrt{7}}{2\\sqrt{3}}$",
"c": "c) $(7\\sqrt{32} + 3\\sqrt{18} - 2\\sqrt{8}) : \\sqrt{2} = \\frac{7\\sqrt{32} + 3\\sqrt{18} - 2\\sqrt{8}}{\\sqrt{2}} = \\frac{7\\sqrt{32}}{\\sqrt{2}} + \\frac{3\\sqrt{18}}{\\sqrt{2}} - \\frac{2\\sqrt{8}}{\\sqrt{2}} = 7\\sqrt{16} + 3\\sqrt{9} - 2\\sqrt{4} = 33$",
"d": "d) $\\frac{\\sqrt{18}}{\\sqrt{2}} - \\frac{\\sqrt{12}}{\\sqrt{3}} = \\frac{\\sqrt{18} \\times \\sqrt{3}}{\\sqrt{2} \\times \\sqrt{3}} - \\frac{\\sqrt{12} \\times \\sqrt{2}}{\\sqrt{3} \\times \\sqrt{2}} = \\frac{\\sqrt{54}}{\\sqrt{6}} - \\frac{\\sqrt{24}}{\\sqrt{6}} = \\frac{3\\sqrt{6}}{\\sqrt{6}} - \\frac{2\\sqrt{6}}{\\sqrt{6}} = 1$"
},
"content4": {
"a": "a) $\\sqrt{50} : \\sqrt{18} = \\frac{\\sqrt{50}}{\\sqrt{18}} = \\frac{\\sqrt{25} \\times \\sqrt{2}}{\\sqrt{9}} = \\frac{5\\sqrt{2}}{3}$",
"b": "b) $\\sqrt{\\frac{7}{6}} : \\sqrt{\\frac{14}{9}} = \\sqrt{\\frac{7}{6}} \\div \\sqrt{\\frac{14}{9}} = \\frac{\\sqrt{7} \\times \\sqrt{9}}{\\sqrt{6} \\times \\sqrt{14}} = \\frac{3\\sqrt{7}}{2\\sqrt{3}}$",
"c": "c) $(7\\sqrt{32} + 3\\sqrt{18} - 2\\sqrt{8}) : \\sqrt{2} = \\frac{7\\sqrt{32} + 3\\sqrt{18} - 2\\sqrt{8}}{\\sqrt{2}} = \\frac{7\\sqrt{2^5} + 3\\sqrt{2^2 \\times 3} - 2\\sqrt{2^3}}{\\sqrt{2}} = \\frac{7\\times 2^4 + 3\\times 2\\sqrt{3} - 2\\times 2\sqrt{2}}{\\sqrt{2}} = 33$",
"d": "d) $\\frac{\\sqrt{18}}{\\sqrt{2}} - \\frac{\\sqrt{12}}{\\sqrt{3}} = \\frac{\\sqrt{18} \\times \\sqrt{3}}{\\sqrt{2} \\times \\sqrt{3}} - \\frac{\\sqrt{12} \\times \\sqrt{2}}{\\sqrt{3} \\times \\sqrt{2}} = \\frac{\\sqrt{6^2 \\times 3} }{ \\sqrt{2^3} } - \\frac{\\sqrt{2^2 \\times 3} }{\\sqrt{3^1} } = 1$"
}
}