与えられた6つの数式を計算して簡単にします。代数学平方根計算展開有理化2025/4/201. 問題の内容与えられた6つの数式を計算して簡単にします。2. 解き方の手順(1) 43+53−734\sqrt{3} + 5\sqrt{3} - 7\sqrt{3}43+53−733\sqrt{3}3でくくると、(4+5−7)3=23(4+5-7)\sqrt{3} = 2\sqrt{3}(4+5−7)3=23(2) 350−418+323\sqrt{50} - 4\sqrt{18} + \sqrt{32}350−418+3250=25×2=52\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}50=25×2=5218=9×2=32\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}18=9×2=3232=16×2=42\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}32=16×2=42これらを代入すると、3(52)−4(32)+42=152−122+42=(15−12+4)2=723(5\sqrt{2}) - 4(3\sqrt{2}) + 4\sqrt{2} = 15\sqrt{2} - 12\sqrt{2} + 4\sqrt{2} = (15-12+4)\sqrt{2} = 7\sqrt{2}3(52)−4(32)+42=152−122+42=(15−12+4)2=72(3) (7+2)(7−2)(\sqrt{7}+2)(\sqrt{7}-2)(7+2)(7−2)これは (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2(a+b)(a−b)=a2−b2 の形なので、(7)2−(2)2=7−4=3(\sqrt{7})^2 - (2)^2 = 7 - 4 = 3(7)2−(2)2=7−4=3(4) (42−33)(52+23)(4\sqrt{2}-3\sqrt{3})(5\sqrt{2}+2\sqrt{3})(42−33)(52+23)展開すると、42×52+42×23−33×52−33×234\sqrt{2} \times 5\sqrt{2} + 4\sqrt{2} \times 2\sqrt{3} - 3\sqrt{3} \times 5\sqrt{2} - 3\sqrt{3} \times 2\sqrt{3}42×52+42×23−33×52−33×23=20×2+86−156−6×3=40+86−156−18=22−76= 20 \times 2 + 8\sqrt{6} - 15\sqrt{6} - 6 \times 3 = 40 + 8\sqrt{6} - 15\sqrt{6} - 18 = 22 - 7\sqrt{6}=20×2+86−156−6×3=40+86−156−18=22−76(5) (3+26)2(\sqrt{3}+2\sqrt{6})^2(3+26)2(3)2+2(3)(26)+(26)2(\sqrt{3})^2 + 2(\sqrt{3})(2\sqrt{6}) + (2\sqrt{6})^2(3)2+2(3)(26)+(26)2=3+418+4×6=3+4×32+24=27+122= 3 + 4\sqrt{18} + 4 \times 6 = 3 + 4 \times 3\sqrt{2} + 24 = 27 + 12\sqrt{2}=3+418+4×6=3+4×32+24=27+122(6) (32−27)2(3\sqrt{2}-2\sqrt{7})^2(32−27)2(32)2−2(32)(27)+(27)2(3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{7}) + (2\sqrt{7})^2(32)2−2(32)(27)+(27)2=9×2−1214+4×7=18−1214+28=46−1214= 9 \times 2 - 12\sqrt{14} + 4 \times 7 = 18 - 12\sqrt{14} + 28 = 46 - 12\sqrt{14}=9×2−1214+4×7=18−1214+28=46−12143. 最終的な答え(1) 232\sqrt{3}23(2) 727\sqrt{2}72(3) 333(4) 22−7622 - 7\sqrt{6}22−76(5) 27+12227 + 12\sqrt{2}27+122(6) 46−121446 - 12\sqrt{14}46−1214