[tex]\sqrt{x^{2}+48} = 4x - 3 + \sqrt{x^{2}+35}[/tex]
Vì [tex]\sqrt{x^{2}+48}>0[/tex] mà [TEX]\sqrt{x^{2}+48} = 4x - 3 + \sqrt{x^{2}+35}[/TEX] nên [tex]4x - 3 + \sqrt{x^{2}+35}>0[/tex]
Mặt khác: [tex]\sqrt{x^{2}+35}>0[/tex]
Suy ra [tex]4x-3>0\Leftrightarrow x>\frac{3}{4}[/tex]
Ta có:
[tex]\sqrt{x^{2}+48} = 4x - 3 + \sqrt{x^{2}+35}\\\Leftrightarrow \sqrt{x^{2}+48}-7 = 4x - 4 + \sqrt{x^{2}+35}-6\\\Leftrightarrow \frac{x^2-1}{ \sqrt{x^{2}+48}+7}=4(x-1)+\frac{x^2-1}{\sqrt{x^{2}+35}+6}=0\\\Leftrightarrow (x-1)\left ( 4+\frac{x+1}{\sqrt{x^2+35}+6}-\frac{x+1}{\sqrt{x^2+48}+7} \right )=0[/tex]
Th1: [tex]x-1=0\Leftrightarrow x=1(t/m)[/tex]
Th2: [tex]4+\frac{x+1}{\sqrt{x^2+35}+6}-\frac{x+1}{\sqrt{x^2+48}+7}=0\\\Leftrightarrow 4=\frac{x+1}{\sqrt{x^2+48}+7}-\frac{x+1}{\sqrt{x^2+35}+6}(*)[/tex]
Vì [tex]x> \frac{3}{4}\Rightarrow x+1>0[/tex]
Ta có [tex]\sqrt{x^2+28}+7>\sqrt{x^2+35}+6\Rightarrow \frac{x+1}{\sqrt{x^2+48}+7}<\frac{x+1}{\sqrt{x^2+35}+6}\Rightarrow \frac{x+1}{\sqrt{x^2+48}+7}-\frac{x+1}{\sqrt{x^2+35}+6}<0<4[/tex]
Suy ra (*) vô nghiệm
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