جواب فعالیت صفحه 65 درس 3 ریاضی دهم (توان های گویا و عبارت های جبری)

\(\begin{array}{l}\frac{5}{{2\sqrt 3 }} = \frac{5}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{5\sqrt 3 }}{6}\\\\\frac{2}{{\sqrt[3]{4}}} = \frac{2}{{\sqrt[3]{4}}} \times \frac{{\sqrt[3]{{{4^2}}}}}{{\sqrt[3]{{{4^2}}}}} = \frac{{2\sqrt[3]{{{4^2}}}}}{4} = \frac{{\sqrt[3]{{{4^2}}}}}{2} = \frac{{\sqrt[3]{{16}}}}{2}\end{array}\)

الف

\(\begin{array}{l}\frac{1}{{\sqrt 5 - \sqrt 3 }} = \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} = \\\\\frac{{\sqrt 5 + \sqrt 3 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} = \frac{{\sqrt 5 + \sqrt 3 }}{{5 - 3}} = \frac{{\sqrt 5 + \sqrt 3 }}{2}\end{array}\)

ب

\(\begin{array}{l}\frac{8}{{3\sqrt 2 + 4}} = \frac{8}{{3\sqrt 2 + 4}} \times \frac{{3\sqrt 2 - 4}}{{3\sqrt 2 - 4}} = \\\\\frac{{8\left( {3\sqrt 2 - 4} \right)}}{{{{\left( {3\sqrt 2 } \right)}^2} - {{\left( 4 \right)}^2}}} = \frac{{24\sqrt 2 - 32}}{{18 - 16}} = \\\\\frac{{24\sqrt 2 - 32}}{2} = 12\sqrt 2 - 16\end{array}\)

پ

\(\begin{array}{l}\frac{{x - y}}{{\sqrt x - \sqrt y }} = \frac{{x - y}}{{\sqrt x - \sqrt y }} \times \frac{{\sqrt x + \sqrt y }}{{\sqrt x + \sqrt y }} = \\\\\frac{{\left( {x - y} \right)\left( {\sqrt x + \sqrt y } \right)}}{{{{\left( {\sqrt x } \right)}^2} - {{\left( {\sqrt y } \right)}^2}}} = \frac{{\left( {x - y} \right)\left( {\sqrt x + \sqrt y } \right)}}{{x - y}} = \sqrt x + \sqrt y \end{array}\)

ت

\(\begin{array}{l}\frac{h}{{\sqrt {x + h} - \sqrt x }} = \\\\\frac{h}{{\sqrt {x + h} - \sqrt x }} \times \frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }} = \\\\\frac{{h\left( {\sqrt {x + h} + \sqrt x } \right)}}{{{{\left( {\sqrt {x + h} } \right)}^2} - {{\left( {\sqrt x } \right)}^2}}} = \frac{{h\left( {\sqrt {x + h} + \sqrt x } \right)}}{{x + h - x}} = \\\\\frac{{h\left( {\sqrt {x + h} + \sqrt x } \right)}}{h} = \sqrt {x + h} + \sqrt x \end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) = {x^3} - {y^3}\\\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)\left( {\sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right) = \;\sqrt[3]{{{a^3}}} - \sqrt[3]{{{b^3}}} = a - b\end{array} \right.\\\\\left\{ \begin{array}{l}\left( {x + y} \right)\left( {{x^2} + xy + {y^2}} \right) = {x^3} + {y^3}\\\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left( {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right) = \;\sqrt[3]{{{a^3}}} + \sqrt[3]{{{b^3}}} = a + b\end{array} \right.\end{array}\)

خیر، سمت راست تساوی های بالا، شامل یک عبارت رادیکالی نیست.

الف

\(\begin{array}{l}\frac{2}{{\sqrt[3]{5} + 1}} = \\\\\frac{2}{{\sqrt[3]{5} + 1}} \times \frac{{\sqrt[3]{{{5^2}}} - \sqrt[3]{5} + 1}}{{\sqrt[3]{{{5^2}}} - \sqrt[3]{5} + 1}} = \\\\\frac{{2\left( {\sqrt[3]{{{5^2}}} - \sqrt[3]{5} + 1} \right)}}{{{{\left( {\sqrt[3]{5}} \right)}^3} + {{\left( 1 \right)}^3}}} = \\\\\frac{{2\left( {\sqrt[3]{{{5^2}}} - \sqrt[3]{5} + 1} \right)}}{{5 + 1}} = \\\\\frac{{\sqrt[3]{{{5^2}}} - \sqrt[3]{5} + 1}}{3}\end{array}\)

ب

\(\begin{array}{l}\frac{1}{{\sqrt[3]{7} - \sqrt[3]{3}}} = \\\\\frac{1}{{\sqrt[3]{7} - \sqrt[3]{3}}} \times \frac{{\sqrt[3]{{{7^2}}} + \sqrt[3]{{21}} + \sqrt[3]{{{3^2}}}}}{{\sqrt[3]{{{7^2}}} + \sqrt[3]{{21}} + \sqrt[3]{{{3^2}}}}} = \\\\\frac{{\sqrt[3]{{{7^2}}} + \sqrt[3]{{21}} + \sqrt[3]{{{3^2}}}}}{{{{\left( {\sqrt[3]{7}} \right)}^3} - {{\left( {\sqrt[3]{3}} \right)}^3}}} = \\\\\frac{{\sqrt[3]{{49}} + \sqrt[3]{{21}} + \sqrt[3]{9}}}{{7 - 3}} = \\\\\frac{{\sqrt[3]{{49}} + \sqrt[3]{{21}} + \sqrt[3]{9}}}{4}\end{array}\)

پ

\(\begin{array}{l}\frac{{x + 8}}{{\sqrt[3]{x} + 2}} = \\\\\frac{{x + 8}}{{\sqrt[3]{x} + 2}} \times \frac{{\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4}}{{\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4}} = \\\\\frac{{\left( {x + 8} \right)\left( {\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4} \right)}}{{{{\left( {\sqrt[3]{x}} \right)}^3} + {{\left( 2 \right)}^3}}} = \\\\\frac{{\left( {x + 8} \right)\left( {\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4} \right)}}{{x + 8}} = \\\\\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4\end{array}\)

ت

\(\begin{array}{l}\frac{1}{{\sqrt[3]{{{x^2}}} + 1}} = \\\\\frac{1}{{\sqrt[3]{{{x^2}}} + 1}} \times \frac{{\sqrt[3]{{{x^4}}} - \sqrt[3]{{{x^2}}} + 1}}{{\sqrt[3]{{{x^4}}} - \sqrt[3]{{{x^2}}} + 1}} = \\\\\frac{{\sqrt[3]{{{x^4}}} - \sqrt[3]{{{x^2}}} + 1}}{{{{\left( {\sqrt[3]{{{x^2}}}} \right)}^3} + {{\left( 1 \right)}^3}}} = \frac{{\sqrt[3]{{{x^4}}} - \sqrt[3]{{{x^2}}} + 1}}{{{x^2} + 1}}\end{array}\)