جواب تمرین صفحه 76 درس 4 ریاضی نهم (توان و ریشه)

الف

\(\begin{array}{l}2\sqrt {50} + \sqrt {32} + 2\sqrt {72} = \\\\2\sqrt {25 \times 2} + \sqrt {16 \times 2} + 2\sqrt {36 \times 2} = \\\\10\sqrt 2 + 4\sqrt 2 + 12\sqrt 2 = 26\sqrt 2 \end{array}\)

ب

\(\begin{array}{l}\sqrt 8 + \sqrt {128} - \sqrt {50} = \\\\\sqrt {4 \times 2} + \sqrt {64 \times 2} - \sqrt {25 \times 2} = \\\\2\sqrt 2 + 8\sqrt 2 - 5\sqrt 2 = 5\sqrt 2 \end{array}\)

ج

\(\begin{array}{l}\sqrt[3]{{{{27}^2}}} = \sqrt[3]{{{{({3^2})}^2}}} = \\\\\sqrt[3]{{{{({3^2})}^3}}} = \sqrt[3]{{{{\left( 9 \right)}^3}}} = 9\end{array}\)

د

\(\begin{array}{l}\sqrt[3]{{\frac{{ - 27}}{{64}}}} = \sqrt[3]{{\frac{{ - {{({3^3})}^2}}}{{{4^3}}}}} = \\\\\sqrt[3]{{{{\left( { - \frac{3}{4}} \right)}^3}}} = - \frac{3}{4}\end{array}\)

هـ

\(\begin{array}{l}\left( {\sqrt 2 - \sqrt 5 } \right)\left( {\sqrt {10} + \sqrt 2 } \right) = \\\\\sqrt {20} + \sqrt 4 - \sqrt {50} - \sqrt {10} = \\\\\sqrt {4 \times 5} + 2 - \sqrt {25 \times 2} - \sqrt {25 \times 2} - \sqrt {10} = \\\\2\sqrt 5 + 2 - 5\sqrt 2 - \sqrt {10} \end{array}\)

و

\(\begin{array}{l}2\sqrt {48} - 3\sqrt {27} = \\\\2\sqrt {16 \times 3} - 3\sqrt {9 \times 3} = \\\\2 \times 4\sqrt 3 - 3 \times 3\sqrt 3 = \\\\8\sqrt 3 - 9\sqrt 3 = - \sqrt 3 \end{array}\)

\(\begin{array}{l}2\sqrt {{x^2}} - x = 2\left| x \right| - x\\\\\mathop \Rightarrow \limits^{x < 0} \,\,\,\,\, - 2x - x = - 3x\end{array}\)

\(s = 3\sqrt 5 \times 3\sqrt 5 = 9\sqrt {25} = 9 \times 5 = 45\)

محیط \( = 2 \times 3\sqrt 5 = 12\sqrt 5 \)

\(\begin{array}{l}A{H^2} = A{C^2} - H{C^2}\\\\ \Rightarrow {h^2} = {a^2} - {\left( {\frac{a}{2}} \right)^2}\\\\ \Rightarrow \;{h^2} = {a^2} - \frac{{{a^2}}}{4} = \frac{{4{a^2} - {a^2}}}{4} = \frac{{3{a^2}}}{4}\\\\ \Rightarrow {h^2} = \frac{{3{a^2}}}{4}\\\\ \Rightarrow {h^2} = \frac{{\sqrt 3 a}}{4}\end{array}\)

\(\begin{array}{l}ABCD{\rm{\;\;}}S = 100\; \Rightarrow \;{\rm{AB}} = \sqrt {100} = 10 \Rightarrow BM = AM = \frac{{10}}{2} = 5\\\\BMN\;{\rm{\;}}:\;\hat B = 90 \Rightarrow M{N^2} = B{M^2} + B{N^2} \Rightarrow M{N^2} = {5^2} + {5^2} \Rightarrow M{N^2} = 50\\\\ \Rightarrow MN = \sqrt {50} \Rightarrow {\rm{MN}} = \sqrt {25 \times 2} \Rightarrow MN = 5\sqrt 2 \Rightarrow MNPQ{\rm{\;}} = 4\left( {5\sqrt 2 } \right) = 20\sqrt 2 \end{array}\)

\(\begin{array}{l}\sqrt 5 + \sqrt 4 > \sqrt {5 + 4} \\\\4 > \sqrt {{3^2} + {2^2}} \\\\\sqrt {\frac{3}{{11}}} = \frac{{\sqrt 3 }}{{\sqrt {11} }}\\\\\sqrt {{3^2} + {4^2}} = 5\end{array}\)

الف\(\sqrt {100} = 10\)

ب\(2\sqrt 9 = 6\)

ج\(\sqrt {\frac{1}{9}} = \frac{1}{3}\)

د\(\sqrt[3]{8} = 2\)

هـ\(\frac{{{2^{ - 5}}}}{{{2^{ - 8}}}} = \sqrt {64} \)

و\(\frac{{{{(\sqrt {12} )}^2}}}{{4 \times {3^2}}} = {3^{ - 1}}\)

ز\(\frac{{{m^6} \times {m^{ - 2}}}}{{{m^3}}} = m\)

ح\(9\sqrt[3]{{ - 27}} = \frac{{{{12}^3}}}{{{{( - 4)}^3}}}\)

الف\(\frac{5}{{2\sqrt 3 }} = \frac{5}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{5 \times \sqrt 3 }}{{2 \times 3}} = \frac{{5\sqrt 3 }}{6}\)

ب\(\frac{2}{{\sqrt[3]{{{a^2}}}}} = \frac{2}{{\sqrt[3]{{{a^2}}}}} \times \frac{{\sqrt[3]{a}}}{{\sqrt[3]{a}}} = \frac{{2\sqrt[3]{a}}}{a}\)

ج\(\frac{2}{{\sqrt 7 }} = \frac{2}{{\sqrt 7 }} \times \frac{{\sqrt 7 }}{{\sqrt 7 }} = \frac{{2\sqrt 7 }}{7}\)

پاسخ صحیح گزینۀ ج می باشد؛ چون:

\(\sqrt {{x^2}} = \left| x \right|\)

و

\({(\sqrt x )^2} = x\)

این دو فقط وقتی برابرند که \(x \ge 0\)  باشد.