[tex]9n^2-4n+12= (n^3+n)(3+\sqrt{5-n^2})\\\\ <=> n^2-4n+4+8n^2+8- 3n^3-3n- (n^3+n).\sqrt{5-n^2}=0\\\\ <=> (n-2)^2+8n^2+8-3n^3-3n-(n^3+n).[\sqrt{5-n^2}-(5-2n)]-(n^3+n)(5-2n)=0\\\\ <=> (n-2)^2+8n^2+8-3n^3-3n-(n^3+n).\frac{-5(n-2)^2}{\sqrt{5-n^2}+5-2n}+(n^3+n)(2n-5)=0\\\\ <=> (n-2)^2+\frac{5(n-2)^2(n^3+n)}{\sqrt{5-n^2}+5-2n}+2n^4+2n^2-5n^3-5n-3n^3-3n+8n^2+8=0\\\\ <=> (n-2)^2+\frac{5(n-2)^2(n^3+n)}{\sqrt{5-n^2}+5-2n}+(2n^4-8n^3+8n^2)+(2n^2-8n+8)=0\\\\ <=> (n-2)^2.[1+\frac{5(n^3+n)}{\sqrt{5-n^2}+5-2n}+2n^2+2]=0[/tex]
từ pt đầu dễ cm: VP>0 suy ra n>0
mặt khác: [tex]n\leq \sqrt{5} => 2n\leq 2\sqrt{5}<5\\\\ => 5-2n>0[/tex]
suy ra pt trong ngoặc dương
suy ra pt <=> n=2
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
