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Demonstrati ca  \frac{1}{x ^{4} } + \frac{1}{y ^{4} } este numar natural
x= 5\sqrt{2} -7
y=5 \sqrt{2} +7
REZOLVARE COMPLETA! URGENT!


Răspuns :

Cpw
[tex] \frac{1}{(5 \sqrt{2}-7)^4} + \frac{1}{(5 \sqrt{2}+7)^4} =[/tex]

[tex] =\frac{(5 \sqrt{2}+7)^4}{(5 \sqrt{2}-7)^4(5 \sqrt{2}+7)^4} + \frac{(5 \sqrt{2}-7)^4}{(5 \sqrt{2}-7)^4(5 \sqrt{2}+7)^4} =[/tex]

[tex] =\frac{(5 \sqrt{2}+7)^4+(5 \sqrt{2}-7)^4}{(5 \sqrt{2}-7)^4(5 \sqrt{2}+7)^4} =[/tex]

[tex]=\frac{(5 \sqrt{2}+7)^4+(5 \sqrt{2}-7)^4}{[(5 \sqrt{2}-7)(5 \sqrt{2}+7)]^4} =[/tex]

[tex]=\frac{(5 \sqrt{2}+7)^4+(5 \sqrt{2}-7)^4}{[(5 \sqrt{2})^2-7^2]^4} =[/tex]

[tex]=\frac{(5 \sqrt{2}+7)^4+(5 \sqrt{2}-7)^4}{(50-49)^4} =\frac{(5 \sqrt{2}+7)^4+(5 \sqrt{2}-7)^4}{1^4} =[/tex]

[tex]={(5 \sqrt{2}+7)^4+(5 \sqrt{2}-7)^4=[/tex]  

=625*4-4*125*2*√2*7+6*25*2*49-4*5*√2*343+2401 +625*4+4*125*2*√2*7+6*25*2*49+4*5*√2*343+2401=

=625*4+6*25*2*49+2401 +625*4+6*25*2*49+2401 =

=2500+14700+2401 +2500+14700+2401=

=39202

este numar natural