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画像に掲載されている数学の問題を解きます。具体的には、以下の問題です。 (1) $12 \times \sqrt{\frac{5}{3}} - \sqrt{270} \div \sqrt{6}$ (2) $(\sqrt{54} + \sqrt{24}) \times \sqrt{2} \div \sqrt{3}$ (3) $\sqrt{3}(\sqrt{2} - \sqrt{6}) - \sqrt{48} \div \sqrt{2} + \frac{6}{\sqrt{2}}$ (4) $(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3}) - (\sqrt{2} - 2)^2$ (5) $(\sqrt{3} + \sqrt{5})^2 - (\sqrt{3} - \sqrt{5})^2$ (6) $(\sqrt{8} + 2)(\sqrt{8} + 3) - \frac{20}{\sqrt{8}}$ (7) $x = \sqrt{3}, y = \sqrt{2}$ のとき、$(x+y)^2 - (x-y)^2$ の値 (8) $x = \sqrt{5} + \sqrt{2}, y = \sqrt{5} - \sqrt{2}$ のとき、$x^2 - y^2$ の値 (9) $x = \sqrt{5} + 2\sqrt{6}, y = \sqrt{5} - 2\sqrt{6}$ のとき、$x^2 + xy + y^2$ の値 (10) $x = \sqrt{3} + \sqrt{2}, y = \sqrt{3} - \sqrt{2}$ のとき、$\frac{y}{x} + \frac{x}{y}$ の値

代数学平方根計算式の展開有理化
2025/8/17

1. 問題の内容

画像に掲載されている数学の問題を解きます。具体的には、以下の問題です。
(1) 12×53−270÷612 \times \sqrt{\frac{5}{3}} - \sqrt{270} \div \sqrt{6}
(2) (54+24)×2÷3(\sqrt{54} + \sqrt{24}) \times \sqrt{2} \div \sqrt{3}
(3) 3(2−6)−48÷2+62\sqrt{3}(\sqrt{2} - \sqrt{6}) - \sqrt{48} \div \sqrt{2} + \frac{6}{\sqrt{2}}
(4) (32−23)(32+23)−(2−2)2(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3}) - (\sqrt{2} - 2)^2
(5) (3+5)2−(3−5)2(\sqrt{3} + \sqrt{5})^2 - (\sqrt{3} - \sqrt{5})^2
(6) (8+2)(8+3)−208(\sqrt{8} + 2)(\sqrt{8} + 3) - \frac{20}{\sqrt{8}}
(7) x=3,y=2x = \sqrt{3}, y = \sqrt{2} のとき、(x+y)2−(x−y)2(x+y)^2 - (x-y)^2 の値
(8) x=5+2,y=5−2x = \sqrt{5} + \sqrt{2}, y = \sqrt{5} - \sqrt{2} のとき、x2−y2x^2 - y^2 の値
(9) x=5+26,y=5−26x = \sqrt{5} + 2\sqrt{6}, y = \sqrt{5} - 2\sqrt{6} のとき、x2+xy+y2x^2 + xy + y^2 の値
(10) x=3+2,y=3−2x = \sqrt{3} + \sqrt{2}, y = \sqrt{3} - \sqrt{2} のとき、yx+xy\frac{y}{x} + \frac{x}{y} の値

2. 解き方の手順

(1)
12×53−270÷6=12×53−2706=12×53−45=12×153−35=415−3512 \times \sqrt{\frac{5}{3}} - \sqrt{270} \div \sqrt{6} = 12 \times \frac{\sqrt{5}}{\sqrt{3}} - \sqrt{\frac{270}{6}} = 12 \times \frac{\sqrt{5}}{\sqrt{3}} - \sqrt{45} = 12 \times \frac{\sqrt{15}}{3} - 3\sqrt{5} = 4\sqrt{15} - 3\sqrt{5}
(2)
(54+24)×2÷3=(36+26)×23=56×23=54=5×2=10(\sqrt{54} + \sqrt{24}) \times \sqrt{2} \div \sqrt{3} = (3\sqrt{6} + 2\sqrt{6}) \times \frac{\sqrt{2}}{\sqrt{3}} = 5\sqrt{6} \times \frac{\sqrt{2}}{\sqrt{3}} = 5\sqrt{4} = 5 \times 2 = 10
(3)
3(2−6)−48÷2+62=6−18−24+622=6−32−26+32=−6\sqrt{3}(\sqrt{2} - \sqrt{6}) - \sqrt{48} \div \sqrt{2} + \frac{6}{\sqrt{2}} = \sqrt{6} - \sqrt{18} - \sqrt{24} + \frac{6\sqrt{2}}{2} = \sqrt{6} - 3\sqrt{2} - 2\sqrt{6} + 3\sqrt{2} = -\sqrt{6}
(4)
(32−23)(32+23)−(2−2)2=(32)2−(23)2−((2)2−42+4)=18−12−(2−42+4)=6−6+42=42(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3}) - (\sqrt{2} - 2)^2 = (3\sqrt{2})^2 - (2\sqrt{3})^2 - ((\sqrt{2})^2 - 4\sqrt{2} + 4) = 18 - 12 - (2 - 4\sqrt{2} + 4) = 6 - 6 + 4\sqrt{2} = 4\sqrt{2}
(5)
(3+5)2−(3−5)2=(32+235+52)−(32−235+52)=(3+215+5)−(3−215+5)=8+215−8+215=415(\sqrt{3} + \sqrt{5})^2 - (\sqrt{3} - \sqrt{5})^2 = (\sqrt{3}^2 + 2\sqrt{3}\sqrt{5} + \sqrt{5}^2) - (\sqrt{3}^2 - 2\sqrt{3}\sqrt{5} + \sqrt{5}^2) = (3 + 2\sqrt{15} + 5) - (3 - 2\sqrt{15} + 5) = 8 + 2\sqrt{15} - 8 + 2\sqrt{15} = 4\sqrt{15}
または、a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b)を使うと、
(3+5)2−(3−5)2=((3+5)+(3−5))((3+5)−(3−5))=(23)(25)=415(\sqrt{3} + \sqrt{5})^2 - (\sqrt{3} - \sqrt{5})^2 = ((\sqrt{3} + \sqrt{5}) + (\sqrt{3} - \sqrt{5}))((\sqrt{3} + \sqrt{5}) - (\sqrt{3} - \sqrt{5})) = (2\sqrt{3})(2\sqrt{5}) = 4\sqrt{15}
(6)
(8+2)(8+3)−208=(22+2)(22+3)−2022=(8+62+42+6)−102=14+102−52=14+52(\sqrt{8} + 2)(\sqrt{8} + 3) - \frac{20}{\sqrt{8}} = (2\sqrt{2} + 2)(2\sqrt{2} + 3) - \frac{20}{2\sqrt{2}} = (8 + 6\sqrt{2} + 4\sqrt{2} + 6) - \frac{10}{\sqrt{2}} = 14 + 10\sqrt{2} - 5\sqrt{2} = 14 + 5\sqrt{2}
(7)
x=3,y=2x = \sqrt{3}, y = \sqrt{2} のとき、
(x+y)2−(x−y)2=x2+2xy+y2−(x2−2xy+y2)=4xy=4×3×2=46(x+y)^2 - (x-y)^2 = x^2 + 2xy + y^2 - (x^2 - 2xy + y^2) = 4xy = 4 \times \sqrt{3} \times \sqrt{2} = 4\sqrt{6}
(8)
x=5+2,y=5−2x = \sqrt{5} + \sqrt{2}, y = \sqrt{5} - \sqrt{2} のとき、
x2−y2=(x+y)(x−y)=((5+2)+(5−2))((5+2)−(5−2))=(25)(22)=410x^2 - y^2 = (x+y)(x-y) = ((\sqrt{5} + \sqrt{2}) + (\sqrt{5} - \sqrt{2}))((\sqrt{5} + \sqrt{2}) - (\sqrt{5} - \sqrt{2})) = (2\sqrt{5})(2\sqrt{2}) = 4\sqrt{10}
(9)
x=5+26,y=5−26x = \sqrt{5} + 2\sqrt{6}, y = \sqrt{5} - 2\sqrt{6} のとき、
x2+xy+y2=(5+26)2+(5+26)(5−26)+(5−26)2=(5+430+24)+(5−24)+(5−430+24)=29+430−19+29−430=39x^2 + xy + y^2 = (\sqrt{5} + 2\sqrt{6})^2 + (\sqrt{5} + 2\sqrt{6})(\sqrt{5} - 2\sqrt{6}) + (\sqrt{5} - 2\sqrt{6})^2 = (5 + 4\sqrt{30} + 24) + (5 - 24) + (5 - 4\sqrt{30} + 24) = 29 + 4\sqrt{30} - 19 + 29 - 4\sqrt{30} = 39
(10)
x=3+2,y=3−2x = \sqrt{3} + \sqrt{2}, y = \sqrt{3} - \sqrt{2} のとき、
yx+xy=y2+x2xy=(3−2)2+(3+2)2(3+2)(3−2)=(3−26+2)+(3+26+2)3−2=5−26+5+261=10\frac{y}{x} + \frac{x}{y} = \frac{y^2 + x^2}{xy} = \frac{(\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{(3 - 2\sqrt{6} + 2) + (3 + 2\sqrt{6} + 2)}{3 - 2} = \frac{5 - 2\sqrt{6} + 5 + 2\sqrt{6}}{1} = 10

3. 最終的な答え

(1) 415−354\sqrt{15} - 3\sqrt{5}
(2) 1010
(3) −6-\sqrt{6}
(4) 424\sqrt{2}
(5) 4154\sqrt{15}
(6) 14+5214 + 5\sqrt{2}
(7) 464\sqrt{6}
(8) 4104\sqrt{10}
(9) 3939
(10) 1010