Odpowiedź:
[tex]n[/tex]
Szczegółowe wyjaśnienie:
[tex]$\frac{n+2+\sqrt{n^2-4}}{n+2-\sqrt{n^2-4}} +\frac{n+2-\sqrt{n^2-4}}{n+2+\sqrt{n^2-4}}=[/tex]
[tex]$=\frac{(n+2+\sqrt{n^2-4})^2+(n+2-\sqrt{n^2-4})^2}{(n+2-\sqrt{n^2-4})(n+2+\sqrt{n^2-4})} =[/tex]
[tex]$=\frac{(n+2)^{2}+2(n+2)\sqrt{n^2-4}+n^{2}-4+(n+2)^{2}-2(n+2)\sqrt{n^2-4}+n^{2}-4}{(n+2)^{2}-(n^{2}-4)} =[/tex]
[tex]$=\frac{2(n+2)^{2}+2(n^{2}-4)}{(n+2)^{2}-n^2+4} =\frac{2n^{2}+8n+8+2n^{2}-8}{n^{2}+4n+4-n^{2}+4} =\frac{4n^{2}+8n}{4n+8} =n[/tex]
można zastąpić (zrobić podstawienie) :
x = n+2 → [tex]x^{2} =(n+2)^{2} =n^{2} +4n+4[/tex]
y = [tex]\sqrt{n^{2}-4 }[/tex] → [tex]y^{2} =(\sqrt{n^{2}-4 })^{2} =n^{2}-4[/tex]
zatem otrzymamy (po sprowadzeniu do wspólnego mianowniku i uproszczeniu) :
SPOSÓB 1.
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} = \frac{x+y}{x-y} +\frac{x-y}{x+y}=\\ \\=\frac{(x+y)^{2} }{(x-y)(x+y)} +\frac{(x-y)^{2} }{(x+y)(x-y)}=\frac{(x+y)^{2}+(x-y)^{2}}{(x+y)(x-y)} =\frac{x^{2} +2xy+y^{2}+x^{2} -2xy+y^{2}}{x^{2} -y^{2} } =\frac{2x^{2} +2y^{2}}{x^{2} -y^{2} } = \\=\frac{2(x^{2} -y^{2})+4y^{2}}{x^{2} -y^{2} } =2+\frac{4y^{2}}{x^{2} -y^{2} }[/tex]
wracamy do podstawieniu:
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} =2+\frac{4y^{2}}{x^{2} -y^{2} }=2+\frac{4n^{2}-16}{n^{2} +4n+4 -n^{2}+4 }=2+\frac{4n^{2}-16}{ 4n+8 }=\\\\=2+\frac{(4n+8)(n-2)}{ 4n+8 }=2+n-2=n[/tex]
lub
SPOSÓB 2.
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} = \frac{x+y}{x-y} +\frac{x-y}{x+y}=\\ \\=\frac{(x+y)^{2} }{(x-y)(x+y)} +\frac{(x-y)^{2} }{(x+y)(x-y)}=\frac{(x+y)^{2}+(x-y)^{2}}{(x+y)(x-y)} =\frac{x^{2} +2xy+y^{2}+x^{2} -2xy+y^{2}}{x^{2} -y^{2} } =\frac{2x^{2} +2y^{2}}{x^{2} -y^{2} }=\frac{2(x^{2} +y^{2})}{x^{2} -y^{2} }[/tex]
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} =\frac{2(x^{2} +y^{2})}{x^{2} -y^{2} }=\frac{2(n^{2} +4n+4 +n^{2} -4)}{n^{2} +4n+4-n^{2}+4}=\\\\=\frac{2( n^{2} +4n)}{ 4n+8}=\frac{n( 4n +8)}{ 4n+8}=n[/tex]
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Verified answer
Odpowiedź:
[tex]n[/tex]
Szczegółowe wyjaśnienie:
[tex]$\frac{n+2+\sqrt{n^2-4}}{n+2-\sqrt{n^2-4}} +\frac{n+2-\sqrt{n^2-4}}{n+2+\sqrt{n^2-4}}=[/tex]
[tex]$=\frac{(n+2+\sqrt{n^2-4})^2+(n+2-\sqrt{n^2-4})^2}{(n+2-\sqrt{n^2-4})(n+2+\sqrt{n^2-4})} =[/tex]
[tex]$=\frac{(n+2)^{2}+2(n+2)\sqrt{n^2-4}+n^{2}-4+(n+2)^{2}-2(n+2)\sqrt{n^2-4}+n^{2}-4}{(n+2)^{2}-(n^{2}-4)} =[/tex]
[tex]$=\frac{2(n+2)^{2}+2(n^{2}-4)}{(n+2)^{2}-n^2+4} =\frac{2n^{2}+8n+8+2n^{2}-8}{n^{2}+4n+4-n^{2}+4} =\frac{4n^{2}+8n}{4n+8} =n[/tex]
Odpowiedź:
[tex]n[/tex]
Szczegółowe wyjaśnienie:
można zastąpić (zrobić podstawienie) :
x = n+2 → [tex]x^{2} =(n+2)^{2} =n^{2} +4n+4[/tex]
y = [tex]\sqrt{n^{2}-4 }[/tex] → [tex]y^{2} =(\sqrt{n^{2}-4 })^{2} =n^{2}-4[/tex]
zatem otrzymamy (po sprowadzeniu do wspólnego mianowniku i uproszczeniu) :
SPOSÓB 1.
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} = \frac{x+y}{x-y} +\frac{x-y}{x+y}=\\ \\=\frac{(x+y)^{2} }{(x-y)(x+y)} +\frac{(x-y)^{2} }{(x+y)(x-y)}=\frac{(x+y)^{2}+(x-y)^{2}}{(x+y)(x-y)} =\frac{x^{2} +2xy+y^{2}+x^{2} -2xy+y^{2}}{x^{2} -y^{2} } =\frac{2x^{2} +2y^{2}}{x^{2} -y^{2} } = \\=\frac{2(x^{2} -y^{2})+4y^{2}}{x^{2} -y^{2} } =2+\frac{4y^{2}}{x^{2} -y^{2} }[/tex]
wracamy do podstawieniu:
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} =2+\frac{4y^{2}}{x^{2} -y^{2} }=2+\frac{4n^{2}-16}{n^{2} +4n+4 -n^{2}+4 }=2+\frac{4n^{2}-16}{ 4n+8 }=\\\\=2+\frac{(4n+8)(n-2)}{ 4n+8 }=2+n-2=n[/tex]
lub
SPOSÓB 2.
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} = \frac{x+y}{x-y} +\frac{x-y}{x+y}=\\ \\=\frac{(x+y)^{2} }{(x-y)(x+y)} +\frac{(x-y)^{2} }{(x+y)(x-y)}=\frac{(x+y)^{2}+(x-y)^{2}}{(x+y)(x-y)} =\frac{x^{2} +2xy+y^{2}+x^{2} -2xy+y^{2}}{x^{2} -y^{2} } =\frac{2x^{2} +2y^{2}}{x^{2} -y^{2} }=\frac{2(x^{2} +y^{2})}{x^{2} -y^{2} }[/tex]
wracamy do podstawieniu:
[tex]\frac{n+2+\sqrt{n^{2}-4 }}{n+2-\sqrt{n^{2}-4 }} +\frac{n+2-\sqrt{n^{2}-4 }}{n+2+\sqrt{n^{2}-4}} =\frac{2(x^{2} +y^{2})}{x^{2} -y^{2} }=\frac{2(n^{2} +4n+4 +n^{2} -4)}{n^{2} +4n+4-n^{2}+4}=\\\\=\frac{2( n^{2} +4n)}{ 4n+8}=\frac{n( 4n +8)}{ 4n+8}=n[/tex]