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1 حاصل هر عبارت را به دست آورید:
\(\begin{array}{l}\sqrt {81} = \\\\\sqrt {{4^2}} = \\\\\sqrt {{{\left( { - 4} \right)}^2}} = \\\\\sqrt[3]{{ - 1}} = \\\\\sqrt[3]{{\frac{{27}}{{125}}}} = \\\\\sqrt[3]{{{6^3}}} = \\\\\sqrt[3]{{ - \frac{8}{{1000}}}} = \;\\\\\sqrt[3]{{{{\left( { - 7} \right)}^3}}} = \end{array}\)
\(\begin{array}{l}\sqrt {81} = 9\\\\\sqrt {{4^2}} = 4\\\\\sqrt {{{\left( { - 4} \right)}^2}} = \left| { - 4} \right| = 4\;\;\;\;\;\;\;\;\;\\\\\sqrt[3]{{ - 1}} = - 1\;\;\\\\\sqrt[3]{{\frac{{27}}{{125}}}} = \frac{3}{5}\;\;\;\;\;\;\;\;\;\;\\\\\sqrt[3]{{{6^3}}} = 6\;\;\;\;\;\;\;\;\;\\\\\sqrt[3]{{ - \frac{8}{{1000}}}} = - \frac{2}{{10}}\;\;\;\;\;\;\;\;\;\\\\\sqrt[3]{{{{\left( { - 7} \right)}^3}}} = - 7\;\end{array}\)
2 به کمک رابطه \(\sqrt {{x^2}} = \left| {\left. x \right|} \right.\) ، که در فصل 2 آموخته اید، حاصل عبارت های زیر را به دست آورید:
\(\begin{array}{l}\sqrt {{{\left( { - 6} \right)}^2}} = \\\\\sqrt {{8^2}} = \\\\\sqrt {{{\left( { - \frac{3}{5}} \right)}^2}} = \\\\\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} = \\\\\sqrt {{{\left( {2 - 9} \right)}^2}} = \\\\\sqrt {{{\left( {1 - \frac{1}{3}} \right)}^2}} = \end{array}\)
\(\begin{array}{l}\sqrt {{{\left( { - 6} \right)}^2}} = \left| { - 6} \right| = 6\\\\\sqrt {{8^2}} = \left| 8 \right| = 8\\\\\sqrt {{{\left( { - \frac{3}{5}} \right)}^2}} = \left| { - \frac{3}{5}} \right| = \frac{3}{5}\\\\\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} = \left| {1 - \sqrt 2 } \right| = \sqrt 2 - 1\\\\\sqrt {{{\left( {2 - 9} \right)}^2}} = \left| {2 - 9} \right| = \left| { - 7} \right| = 7\;\\\\\sqrt {{{\left( {1 - \frac{1}{3}} \right)}^2}} = \left| {1 - \frac{1}{3}} \right| = 1 - \frac{1}{3} = \frac{2}{3}\end{array}\)
3 حاصل عبارت \(\sqrt {{x^2}} + \sqrt {{y^2}} \) را در هریک از حالت های زیر بدست آورید؛ یکی از حالت ها حل شده است.
الف x و y هر دو مثبت هستند (x>0 , y>0)
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \)
ب x مثبت و y منفی است (x>0 , y<0)
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \left| {\left. x \right|} \right. + \left| {\left. y \right|} \right. = x - y\)
ج x منفی و y مثبت است (x<0 , y>0)
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \)
د x و y هردو منفی هستند (x<0 , y<0)
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \)
الف
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \left| x \right| + \left| y \right| = x + y\)
ج
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \left| x \right| + \left| y \right| = - x + y\)
د
\(\sqrt {{x^2}} + \sqrt {{y^2}} = \left| x \right| + \left| y \right| = - x - y\)
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