جواب تمرین صفحه 67 درس 3 ریاضی دهم (توان های گویا و عبارت های جبری)

الف

\(\begin{array}{l}{x^4} - {y^4} = \left( {{x^2} - {y^2}} \right)\left( {{x^2} + {y^2}} \right) = \\\\\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)\end{array}\)

ب

\(\begin{array}{l}{x^6} - {y^6} = \left( {{x^3} - {y^3}} \right)\left( {{x^3} + {y^3}} \right) = \\\\\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\end{array}\)

پ

\(\begin{array}{l}8{a^3} + 27 = {\left( {2a} \right)^3} + {3^3} = \\\\\left( {2a + 3} \right)\left( {{{\left( {2a} \right)}^2} - 3\left( {2a} \right) + {3^2}} \right) = \\\\\left( {2a + 3} \right)\left( {4{a^2} - 6a + 9} \right)\end{array}\)

ت

\(\begin{array}{l}{a^3}{b^6} - 8 = {\left( {a{b^2}} \right)^3} - {2^3} = \\\\\left( {a{b^2} - 2} \right)\left( {{a^2}{b^4} + 2a{b^2} + 4} \right)\end{array}\)

الف

\(\begin{array}{l}\frac{3}{{3 + \sqrt 7 }} = \frac{3}{{3 + \sqrt 7 }} \times \frac{{3 - \sqrt 7 }}{{3 - \sqrt 7 }} = \\\\\frac{{3\left( {3 - \sqrt 7 } \right)}}{{{3^2} - {{\left( {\sqrt 7 } \right)}^2}}} = \frac{{9 - 3\sqrt 7 }}{{9 - 7}} = \frac{{9 - 3\sqrt 7 }}{2}\end{array}\)

ب

\(\begin{array}{l}\frac{8}{{\sqrt 5 + \sqrt 3 }} = \frac{8}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} = \\\\\frac{{8\left( {\sqrt 5 - \sqrt 3 } \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} = \frac{{8\left( {\sqrt 5 - \sqrt 3 } \right)}}{{5 - 3}} = \\\\\frac{{8\left( {\sqrt 5 - \sqrt 3 } \right)}}{2} = 4\left( {\sqrt 5 - \sqrt 3 } \right) = \\\\4\sqrt 5 - 4\sqrt 3 \end{array}\)

پ

\(\begin{array}{l}\frac{1}{{\sqrt[3]{x} - 2}} = \frac{1}{{\sqrt[3]{x} - 2}} \times \frac{{\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4}}{{\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4}} = \\\\\frac{{\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4}}{{{{\left( {\sqrt[3]{x}} \right)}^3} - {2^3}}} = \frac{{\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4}}{{x - 8}}\end{array}\)

ت

\(\begin{array}{l}\frac{6}{{2\sqrt[3]{2} - 1}} = \frac{6}{{2\sqrt[3]{2} - 1}} \times \frac{{{{\left( {2\sqrt[3]{2}} \right)}^2} + 2\sqrt[3]{2} + 1}}{{{{\left( {2\sqrt[3]{2}} \right)}^2} + 2\sqrt[3]{2} + 1}} = \\\\\frac{{6\left( {{{\left( {2\sqrt[3]{2}} \right)}^2} + 2\sqrt[3]{2} + 1} \right)}}{{{{\left( {2\sqrt[3]{2}} \right)}^3} - {1^3}}} = \\\\\frac{{6\left( {{{\left( {2\sqrt[3]{2}} \right)}^2} + 2\sqrt[3]{2} + 1} \right)}}{{15}} = \\\\\frac{{2\left( {{{\left( {2\sqrt[3]{2}} \right)}^2} + 2\sqrt[3]{2} + 1} \right)}}{5} = \\\\\frac{{8\sqrt[3]{4} + 4\sqrt[3]{2} + 2}}{5}\end{array}\)

ب

\(\begin{array}{l}{105^2} = {\left( {100 + 5} \right)^2} = \\\\{100^2} + 2 \times 100 \times 5 + {5^2} = \\\\10000 + 1000 + 25 = 11025\end{array}\)

پ

\(\begin{array}{l}{9999^2} = {\left( {10000 - 1} \right)^2} = \\\\{10000^2} - 2 \times 10000 \times 1 + {1^2} = \\\\100,000,000 - 20,000 + 1 = \\\\99,980,001\end{array}\)

ت

\(\begin{array}{l}{105^3} = {\left( {100 + 5} \right)^3} = \\\\{100^3} + 3 \times {100^2} \times 5 + 3 \times 100 \times {5^2} + {5^3} = \\\\1,000,000 + 150,000 + 25\:{\mkern 1mu} + 7500 + 125 = \\\\1,157,625\end{array}\)

الف

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

ب

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

\(\begin{array}{l}\sqrt {x + 2} - \sqrt {x - 4} = \\\\\sqrt {x + 2} - \sqrt {x - 4} \times \frac{{\sqrt {x + 2} + \sqrt {x - 4} }}{{\sqrt {x + 2} + \sqrt {x - 4} }} = \\\\\frac{{{{(\sqrt {x + 2} )}^2} - {{(\sqrt {x - 4} )}^2}}}{{\sqrt {x + 2} + \sqrt {x - 4} }} = \\\\\frac{{(x + 2) - (x - 4)}}{{\sqrt {x + 2} + \sqrt {x - 4} }} = \\\\\frac{6}{{\sqrt {x + 2} + \sqrt {x - 4} }} = \frac{6}{3} = 2\end{array}\)