[tex]\sqrt{x^2+16}-2\sqrt{x^2-3x+4}=\sqrt{x+1}-1[/tex]
[tex]\sqrt{x^2+16}-2\sqrt{x^2-3x+4}=\sqrt{x+1}-1\\\\ <=> \sqrt{x^2+16}-(\frac{1}{3}x+4)+\sqrt{x^2-3x+4}.(\sqrt{x^2-3x+4}-2)=\sqrt{x+1}-(\frac{1}{3}x+1)+x^2-3x\\\\ <=> \frac{x^2+16-(\frac{1}{3}x+4)^2}{\sqrt{x^2+16}+\frac{1}{3}x+4}+\sqrt{x^2-3x+4}.\frac{x^2-3x+4-4}{\sqrt{x^2-3x+4}+2}=\frac{x+1-(\frac{1}{3}x+1)^2}{\sqrt{x+1}+\frac{1}{3}x+1}+x.(x-3)\\\\ <=> \frac{\frac{8}{9}x^2-\frac{8}{3}x}{\sqrt{x^2+16}+\frac{1}{3}x+4}+\sqrt{x^2-3x+4}.\frac{x^2-3x}{\sqrt{x^2-3x+4}+2}=\frac{\frac{-1}{9}x^2+\frac{1}{3}x}{\sqrt{x+1}+\frac{1}{3}x+1}+x^2-3x=0\\\\ <=> x(x-3).[\frac{8}{9.(\sqrt{x^2+16}+\frac{1}{3}x+4)}+\frac{\sqrt{x^2-3x+4}}{\sqrt{x^2-3x+4}+2}+\frac{1}{9.(\sqrt{x+1}+\frac{1}{3}x+1)}-1]=0\\\\ +, x.(x-3)=0 <=>...\\\\ +,\frac{8}{9.(\sqrt{x^2+16}+\frac{1}{3}x+4)}+\frac{\sqrt{x^2-3x+4}}{\sqrt{x^2-3x+4}+2}+\frac{1}{9.(\sqrt{x+1}+\frac{1}{3}x+1)}-1=0\\\\ <=> \frac{8}{9.(\sqrt{x^2+16}+\frac{1}{3}x+4)}+\frac{1}{9.(\sqrt{x+1}+\frac{1}{3}x+1)}-\frac{2}{\sqrt{x^2-3x+4}+2}=0 (1)[/tex]
[tex]*, \sqrt{x^2-3x+4}+2< 9\sqrt{x+1}+3x+9\\\\ <=> \sqrt{x^2-3x+4}< 3x+7 < 9\sqrt{x+1}+3x+7\\\\ <=> x^2-3x+4< 9x^2+49+42x\\\\ <=> 8x^2+45x+45>0 (luôn đúng)[/tex]
[tex]*, 8\sqrt{x^2-3x+4}+16<9\sqrt{x^2+16}+3x+36\\\\ <=> 8\sqrt{x^2-3x+4}<8\sqrt{x^2+7}<9\sqrt{x^2+16}<9\sqrt{x^2+16}+3x+20[/tex]
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