Question

Ma poate ajuta cineva si pe mine prin a imi explica cum se rezolva genul acesta de exercitii(cu tot cu rezolvare,si cu raspuns)
3√150(6√98-8√50-7√108+10√48)+4√54(3√192-8√72-4√75+11√32)

Answer 1

3√150(6√98-8√50-7√108+10√48)+4√54(3√192-8√72-4√75+11√32)
descompui mai intai ce este sub radical si apoi scoti de sub radical

spre exemplu:
$\sqrt{150}= \sqrt{2*3*5^{2} } =5 \sqrt{6}$
$\sqrt{98}= \sqrt{2*7*7} =7 \sqrt{2}$
$\sqrt{50}= \sqrt{2*5*5}=5 \sqrt{2}$
$\sqrt{108}= \sqrt{2*2*3*3*3}= 6\sqrt{3}$
$\sqrt{48}= \sqrt{6*8}= \sqrt{2*3*2^{3} } =4 \sqrt{3}$
$\sqrt{54}= \sqrt{2* 3^{3}}= 3\sqrt{6}$
$\sqrt{192}= \sqrt{2*96} = \sqrt{2*2*48}= \sqrt{2*2*2*3*2^{3}}=8 \sqrt{3}$
$\sqrt{72}= \sqrt{2*36}= \sqrt{2*2*3*2*3}=6 \sqrt{2}$
$\sqrt{75}=5 \sqrt{3}$
$\sqrt{32}= \sqrt{8*4}= \sqrt{ 2^{3}*2^{2}}=4 \sqrt{2}$

$3*5 \sqrt{6}(6*7 \sqrt{2}-8*5 \sqrt{2}-7*6 \sqrt{3}+10*4 \sqrt{3})+ \\ \\ +4*3 \sqrt{6}(3*8 \sqrt{3}-8*6 \sqrt{2}-4*5 \sqrt{3}+11*4 \sqrt{2}=$
$=15 \sqrt{6} (42 \sqrt{2} -40 \sqrt{2}-42 \sqrt{2}+40 \sqrt{3})+12 \sqrt{6}(24 \sqrt{3}-48 \sqrt{2}-20 \sqrt{3}+$
$+44 \sqrt{2}= \\ \\=30 \sqrt{12}-30 \sqrt{18}+48 \sqrt{6}-48 \sqrt{2}$