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

الف

\(\begin{array}{l}\frac{{{x^6} + 1}}{{{x^4} + 2{x^2} + 1}} = \frac{{{{\left( {{x^2}} \right)}^3} + {1^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \\\\\frac{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{{x^4} - {x^2} + 1}}{{{x^2} + 1}}\end{array}\)

ب

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

پ

\(\frac{{{x^2} + 1}}{{{x^4} - 1}} = \frac{{{x^2} + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}} = \frac{1}{{{x^2} - 1}}\)

ت

\(\begin{array}{l}\frac{{{y^5} - {y^3} - 12y}}{{8{y^2} + 16y}} = \frac{{y\left( {{y^4} - {y^2} - 12} \right)}}{{8y\left( {y + 2} \right)}} = \frac{{y\left( {{y^2} - 4} \right)\left( {{y^2} + 3} \right)}}{{8y\left( {y + 2} \right)}} = \\\\\frac{{y\left( {y + 2} \right)\left( {y - 2} \right)\left( {{y^2} + 3} \right)}}{{8y\left( {y + 2} \right)}} = \frac{{\left( {y - 2} \right)\left( {{y^2} + 3} \right)}}{8} = \frac{{{y^3} - 2{y^2} + 3y - 6}}{8}\end{array}\)

\({\left( {\sqrt[3]{{{x^2}}}} \right)^3} + 1 = \left( {\sqrt[3]{{{x^2}}} + 1} \right)\left( {\sqrt[3]{{{x^4}}} - \sqrt[3]{{{x^2}}} + 1} \right)\)