Exercicis de radicals III

Es pretén exercitar la utilització dels radicals sobre tot fórmules molt concretes, a la pràctica consisteix en agrupar arrels com a una sola arrel completant el radicant i efectuar les arrels tant com sigui possible i preparats per no fer servir la calculadora.

A les solucions es troben diverses estratègies o alternatives acceptables però marcaré amb un asterisc, *, les que van bé per practicar encara que sigui la pitjor forma de fer-ho.

${\sqrt {\;a^{2}\;}}=a$	${\sqrt {\;a^{2}\cdot \;b^{2}\cdot \;c^{2}\;}}=a\cdot b\cdot c$	${\sqrt {\;a^{n}\;}}=a^{n\div 2}$
${\sqrt {\;a\;}}^{\,2}=a$
${\sqrt {\;a\cdot b\;}}={\sqrt {a}}\cdot {\sqrt {b}}$

Exercicis

Productes de dos arrels, s'ha d'utilitzar les fórmules anteriors:

$1){\sqrt {2}}\cdot {\sqrt {2}}=$ Alternatives: $a*)={\sqrt {2\cdot 2}}$ $={\sqrt {2^{2}}}$ = 2. $b)={\sqrt {2\cdot 2}}$ $={\sqrt {4}}$ = 2. $c)={\sqrt {2}}^{\;2}$ = 2.	$2){\sqrt {3}}\cdot {\sqrt {3}}=$ Alternatives: $a*)={\sqrt {3\cdot 3}}$ $={\sqrt {3^{2}}}$ = 3. $b)={\sqrt {3\cdot 3}}$ $={\sqrt {9}}$ = 3. $c)={\sqrt {3}}^{\;2}$ = 3.	$3){\sqrt {8}}\cdot {\sqrt {2}}=$ Alternatives: $a*)={\sqrt {8\cdot 2}}$ $={\sqrt {2^{3}\cdot 2}}$ $={\sqrt {2^{4}}}$ $={\sqrt {(2^{2})^{2}}}$ $=2^{2}.$ $b)={\sqrt {8\cdot 2}}$ $={\sqrt {16}}$ = 4.
$4){\sqrt {27}}\cdot {\sqrt {3}}=$ Alternatives: $a*)={\sqrt {27\cdot 3}}$ $={\sqrt {3^{3}\cdot 3}}$ $={\sqrt {3^{4}}}$ $={\sqrt {(3^{2})^{2}}}$ $=3^{2}.$ $b)={\sqrt {27\cdot 3}}$ $={\sqrt {81}}$ = 9.	$5){\sqrt {10}}\cdot {\sqrt {40}}=$ Alternatives: $a*)={\sqrt {2\cdot 5\cdot 2^{3}\cdot 5}}$ $={\sqrt {2^{4}\cdot 5^{2}}}$ $={\sqrt {(2^{2})^{2}}}\cdot {\sqrt {5^{2}}}$ $=2^{2}\cdot 5.$ $b)={\sqrt {10\cdot 4\cdot 10}}$ $={\sqrt {10^{2}\cdot 4}}$ $=10\cdot 2=20.$ $c)={\sqrt {10\cdot 4\cdot 10}}$ $={\sqrt {400}}$ = 20.	$6){\sqrt {32}}\cdot {\sqrt {2}}=$
$7){\sqrt {8}}\cdot {\sqrt {50}}=$	$8){\sqrt {18}}\cdot {\sqrt {2}}=$	$9){\sqrt {2}}\cdot {\sqrt {98}}=$
$10){\sqrt {27}}\cdot {\sqrt {75}}=$

Producte de tres arrels, s'ha d'utilitzar les fórmules anteriors:

$1){\sqrt {6}}\cdot {\sqrt {21}}\cdot {\sqrt {14}}=$	$2){\sqrt {2}}\cdot {\sqrt {6}}\cdot {\sqrt {3}}=$ Hi ha dues possibilitats: a) ${\sqrt {2}}\cdot {\sqrt {6}}\cdot {\sqrt {3}}$ $={\sqrt {2\cdot 6\cdot 3}}$ $={\sqrt {36}}$ $=6.$ b) ${\sqrt {2}}\cdot {\sqrt {6}}\cdot {\sqrt {3}}$ $={\sqrt {2\cdot 2\cdot 3\cdot 3}}$ $={\sqrt {2^{2}\cdot 3^{2}}}$ $={\sqrt {2^{2}}}\cdot {\sqrt {3^{2}}}$ $=2\cdot 3$ $=6.$
$3){\sqrt {15}}\cdot {\sqrt {12}}\cdot {\sqrt {20}}=$	$4){\sqrt {26}}\cdot {\sqrt {130}}\cdot {\sqrt {20}}=$
$5){\sqrt {8}}\cdot {\sqrt {6}}\cdot {\sqrt {3}}=$ Alternatives: a*) ${\sqrt {8\cdot 6\cdot 3}}$ $={\sqrt {2^{3}\cdot 3\cdot 2\cdot 3}}$ $={\sqrt {2^{4}\cdot 3^{2}}}$ $={\sqrt {2^{4}}}\cdot {\sqrt {3^{2}}}$ $=2^{2}\cdot 3=12.$ b) ${\sqrt {8\cdot 6\cdot 3}}$ $={\sqrt {8\cdot 3\cdot 2\cdot 3}}$ $={\sqrt {16\cdot 9}}$ $={\sqrt {16}}\cdot {\sqrt {9}}$ $=4\cdot 3=12.$	$6){\sqrt {27}}\cdot {\sqrt {6}}\cdot {\sqrt {2}}=$
$7){\sqrt {10}}\cdot {\sqrt {20}}\cdot {\sqrt {2}}=$	$8){\sqrt {35}}\cdot {\sqrt {15}}\cdot {\sqrt {21}}=$
$9){\sqrt {21}}\cdot {\sqrt {27}}\cdot {\sqrt {7}}=$	$10){\sqrt {10}}\cdot {\sqrt {15}}\cdot {\sqrt {6}}=$

S'ha de buidar les arrels dels quadrats ràpidament:

${\sqrt {a^{13}}}$ com que 13 és imparell li traiem una a:

${\sqrt {a^{12}\cdot a}}={\sqrt {a^{12}}}\cdot {\sqrt {a}}=a^{12\div 2}\cdot {\sqrt {a}}=a^{6}\cdot {\sqrt {a}}$

$1){\sqrt {2^{5}}}=$	$2){\sqrt {3^{7}}}=$	$3){\sqrt {5^{5}}}=$
$4){\sqrt {4^{11}}}=$	$5){\sqrt {2^{13}}}=$	$6){\sqrt {3^{9}}}=$
$7){\sqrt {11^{3}}}=$	$8){\sqrt {5^{7}}}=$	$9){\sqrt {7^{15}}}=$