数学演習/中学校3年生/平方根/解答

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平方根(1)[編集]

${\sqrt {36-11}}={\sqrt {25}}={\sqrt {5^{2}}}=5$
${\sqrt {121}}={\sqrt {11^{2}}}=11$
${\sqrt {\frac {169}{49}}}={\sqrt {\frac {13^{2}}{7^{2}}}}={\frac {\sqrt {13^{2}}}{\sqrt {7^{2}}}}={\frac {13}{7}}$
${\sqrt {x^{4}}}={\sqrt {(x^{2})^{2}}}=x^{2}$
${\sqrt {\frac {x^{2}}{y^{2}}}}={\frac {\sqrt {x^{2}}}{\sqrt {y^{2}}}}={\frac {x}{y}}$
${\sqrt {\frac {4x^{4}}{225y^{2}z^{8}}}}={\frac {\sqrt {2^{2}\times (x^{2})^{2}}}{\sqrt {15^{2}\times y^{2}\times (z^{4})^{2}}}}={\frac {2x^{2}}{15yz^{4}}}$
${\sqrt {36x^{2}+96xy+64y^{2}}}={\sqrt {(6x)^{2}+2(6x\times 8y)+(8y)^{2}}}={\sqrt {(6x+8y)^{2}}}=6x+8y$

7.は乗法公式 $a^{2}+2ab+b^{2}=(a+b)^{2}$ に $a=6x,b=8y$ を代入した形になっている。（実際に計算してみよう）

${\sqrt {12}}={\sqrt {3\times 4}}={\sqrt {3\times 2^{2}}}=2{\sqrt {3}}$
${\sqrt {75}}={\sqrt {3\times 25}}={\sqrt {3\times 5^{2}}}=5{\sqrt {3}}$
${\sqrt {396}}={\sqrt {36\times 11}}={\sqrt {6^{2}\times 11}}=6{\sqrt {11}}$
${\sqrt {(x+y)^{3}}}={\sqrt {(x+y)\times (x+y)^{2}}}=(x+y){\sqrt {x+y}}$
$6{\sqrt {396}}+{\sqrt {1331}}=6{\sqrt {36\times 11}}+{\sqrt {121\times 11}}=(36+11){\sqrt {11}}=47{\sqrt {11}}$

${\sqrt {2}}+2{\sqrt {2}}=(1+2){\sqrt {2}}=3{\sqrt {2}}$
${\sqrt {5}}-5{\sqrt {5}}=(1-5){\sqrt {5}}=-4{\sqrt {5}}$
$7+2{\sqrt {2}}+3{\sqrt {3}}+6{\sqrt {12}}=7+2{\sqrt {2}}+3{\sqrt {3}}+6{\sqrt {3\times 2^{2}}}=7+2{\sqrt {2}}+3{\sqrt {3}}+12{\sqrt {3}}=7+2{\sqrt {2}}+15{\sqrt {3}}$
${\sqrt {3}}\times 3{\sqrt {7}}=3{\sqrt {3\times 7}}=3{\sqrt {21}}$
${\sqrt {6}}\times 2{\sqrt {3}}={\sqrt {2\times 3}}\times 2{\sqrt {3}}=2{\sqrt {2\times 3\times 3}}=2{\sqrt {2\times 3^{2}}}=6{\sqrt {2}}$
$({\sqrt {2}}+2)(1+{\sqrt {3}})={\sqrt {2}}+{\sqrt {2\times 3}}+2+2{\sqrt {3}}=2+2{\sqrt {3}}+{\sqrt {6}}+{\sqrt {2}}$
$({\sqrt {5}}-2)^{2}=({\sqrt {5}})^{2}-2\times {\sqrt {5}}\times 2+2^{2}=9-4{\sqrt {5}}$
$({\sqrt {7}}+{\sqrt {3}})({\sqrt {7}}-{\sqrt {3}})=({\sqrt {7}})^{2}-({\sqrt {3}})^{2}=7-3=4$
$(6+{\sqrt {55}})(6-{\sqrt {55}})=6^{2}-({\sqrt {55}})^{2}=36-55=-19$

${\frac {2}{\sqrt {3}}}={\frac {2}{\sqrt {3}}}\times 1={\frac {2}{\sqrt {3}}}\times {\frac {\sqrt {3}}{\sqrt {3}}}={\frac {2{\sqrt {3}}}{\sqrt {3^{2}}}}={\frac {2{\sqrt {3}}}{3}}$
${\frac {\sqrt {6}}{\sqrt {5}}}={\frac {\sqrt {6}}{\sqrt {5}}}\times 1={\frac {\sqrt {6}}{\sqrt {5}}}\times {\frac {\sqrt {5}}{\sqrt {5}}}={\frac {\sqrt {6\times 5}}{\sqrt {5^{2}}}}={\frac {\sqrt {30}}{5}}$
${\frac {3+{\sqrt {2}}}{\sqrt {3}}}={\frac {3+{\sqrt {2}}}{\sqrt {3}}}\times 1={\frac {3+{\sqrt {2}}}{\sqrt {3}}}\times {\frac {\sqrt {3}}{\sqrt {3}}}={\frac {{\sqrt {3}}(3+{\sqrt {2}})}{\sqrt {3^{2}}}}={\frac {3{\sqrt {3}}+{\sqrt {2\times 3}}}{3}}={\frac {3{\sqrt {3}}+{\sqrt {6}}}{3}}$
${\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}={\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}\times 1={\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}\times {\frac {{\sqrt {2}}+1}{{\sqrt {2}}+1}}={\frac {({\sqrt {2}}+1)^{2}}{({\sqrt {2}}-1)({\sqrt {2}}+1)}}={\frac {({\sqrt {2}})^{2}+2\times {\sqrt {2}}\times 1+1^{2}}{({\sqrt {2}})^{2}-1^{2}}}={\frac {3+2{\sqrt {2}}}{1}}=3+2{\sqrt {2}}$