画像にある8つの計算問題を解き、空欄を埋めます。代数学平方根根号計算計算2025/6/211. 問題の内容画像にある8つの計算問題を解き、空欄を埋めます。2. 解き方の手順(1) 3×7=3×7=21\sqrt{3} \times \sqrt{7} = \sqrt{3 \times 7} = \sqrt{21}3×7=3×7=21(2) 24+6=4×6+6=26+6=36\sqrt{24} + \sqrt{6} = \sqrt{4 \times 6} + \sqrt{6} = 2\sqrt{6} + \sqrt{6} = 3\sqrt{6}24+6=4×6+6=26+6=36(3) 52+22=725\sqrt{2} + 2\sqrt{2} = 7\sqrt{2}52+22=72(4) 96−24=16×6−4×6=46−26=26\sqrt{96} - \sqrt{24} = \sqrt{16 \times 6} - \sqrt{4 \times 6} = 4\sqrt{6} - 2\sqrt{6} = 2\sqrt{6}96−24=16×6−4×6=46−26=26(5) 18−22+32=9×2−22+16×2=32−22+42=(3−2+4)2=52\sqrt{18} - 2\sqrt{2} + \sqrt{32} = \sqrt{9 \times 2} - 2\sqrt{2} + \sqrt{16 \times 2} = 3\sqrt{2} - 2\sqrt{2} + 4\sqrt{2} = (3-2+4)\sqrt{2} = 5\sqrt{2}18−22+32=9×2−22+16×2=32−22+42=(3−2+4)2=52(6) 2(5+7)=2×5+2×7=10+14\sqrt{2}(\sqrt{5}+\sqrt{7}) = \sqrt{2} \times \sqrt{5} + \sqrt{2} \times \sqrt{7} = \sqrt{10} + \sqrt{14}2(5+7)=2×5+2×7=10+14(7) (3+5)2=(3)2+235+(5)2=3+215+5=8+215(\sqrt{3}+\sqrt{5})^2 = (\sqrt{3})^2 + 2\sqrt{3}\sqrt{5} + (\sqrt{5})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}(3+5)2=(3)2+235+(5)2=3+215+5=8+215(8) (10−3)(10+3)=(10)2−(3)2=10−3=7(\sqrt{10}-\sqrt{3})(\sqrt{10}+\sqrt{3}) = (\sqrt{10})^2 - (\sqrt{3})^2 = 10 - 3 = 7(10−3)(10+3)=(10)2−(3)2=10−3=73. 最終的な答え(1) 21\sqrt{21}21(2) 363\sqrt{6}36(3) 727\sqrt{2}72(4) 262\sqrt{6}26(5) 525\sqrt{2}52(6) 10+14\sqrt{10} + \sqrt{14}10+14(7) 8+2158 + 2\sqrt{15}8+215(8) 777