Question

efectua la siguiente adicion y sustraccion de radicales
me ayudan pliss
√72 × -√32 ×+ + √128 ×=


3 √2 + 5 √2 - 7 √2 + 4 √2=


6 √2 - 2 √2 + 4 √2 - 5 √2=



3 √2 - 4 √8 + 5 √50 - 3 √32=



2 √3 - 3 √3 + 5 √3 - 4 √3=



2 √5 + 7 √5 - 3 √5 + 8 √5=


4 √12 - 3 √75 + 6 √300 - √108=

Answer 1

Respuesta:

$\sqrt{72}-\sqrt{32}+\sqrt{128}=\\=\sqrt{2^3\cdot \:3^2}-\sqrt{2^4\cdot \:2}+\sqrt{2^6\cdot \:2}=\\=2\cdot3\sqrt{2}-2^{2} \sqrt{2}+2^{3} \sqrt{2}=\\=6\sqrt{2}-4\sqrt{2}+8\sqrt{2}=\\=14\sqrt{2}-4\sqrt{2}=\\=10\sqrt{2}$

$3\:\sqrt{2}\:+\:5\:\sqrt{2}\:-\:7\:\sqrt{2}\:+\:4\:\sqrt{2}=5\sqrt{2}$

$6\:\sqrt{2}\:-\:2\:\sqrt{2}\:+\:4\:\sqrt{2}\:-\:5\:\sqrt{2}=3\sqrt{2}$

$3\:\sqrt{2}\:-\:4\:\sqrt{8}\:+\:5\:\sqrt{5}0\:-\:3\:\sqrt{3}2=3\:\sqrt{2}\:-4\sqrt{2^3}+5\sqrt{5^2\cdot \:2}-3\sqrt{2^4\cdot \:2}=3\:\sqrt{2}\:-4\cdot2\sqrt{2}+5\cdot5\sqrt{2}-3\cdot 2^{2} \sqrt{2}=3\:\sqrt{2}\:-8\sqrt{2}+25\sqrt{2}-12\sqrt{2}=28\:\sqrt{2}\:-20\sqrt{2}=8\:\sqrt{2}$

$2\:\sqrt{3}\:-\:3\:\sqrt{3}\:+\:5\:\sqrt{3}\:-\:4\:\sqrt{3}=7\:\sqrt{3}\:-\:7\:\sqrt{3}\:=0$

$2\:\sqrt{5}\:+\:7\:\sqrt{5}\:-\:3\:\sqrt{5}\:+\:8\:\sqrt{5}=14\sqrt{5}$

$4\:\sqrt{12}-3\:\sqrt{75}+6\:\sqrt{300}-\sqrt{108}=\\=4\sqrt{2^2\cdot \:3}-3\sqrt{5^2\cdot \:3}+6\sqrt{2^2\cdot \:5^2\cdot \:3}-\sqrt{3^3\cdot \:2^2}=\\=4\cdot2\sqrt{3}-3\cdot5\sqrt{3}+6\cdot2\cdot5\sqrt{3}-3\cdot2\sqrt{3}=\\=8\sqrt{3}-15\sqrt{3}+60\sqrt{3}-6\sqrt{3}=\\=68\sqrt{3}-21\sqrt{3}=47\sqrt{3}$