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TXĐBài 27 : [TEX]log_{2008}\frac{4x^2+2}{x^6+x^2+1}=x^6-3x^2-1[/TEX]
[TEX]PT log_{2008}(4x^2+2)-log_{2008}(x^6+x^2+1)=(x^6+x^2+1)-(4x^2+2)[/TEX]
[TEX]\Leftrightarrow log_{2008}(4x^2+2)+(4x^2+2)=log_{2008}(x^6+x^2+1)+(x^6+x^2+1)[/TEX]
[TEX]\Leftrightarrow g(4x^2+2)=g(x^6+x^2+1)[/TEX]
Với [TEX]g(t)=log_{2008}t+t,t > 0[/TEX]
[TEX]g'(t)=\frac{1}{tln2008}+1 > 0 \forall t > 0[/TEX]
\Rightarrow Hàm số đồng biến trên (0;+\infty)
[TEX]\Rightarrow PT \Leftrightarrow 4x^2+2=x^6+x^2+1 \Leftrightarrow x^6-3x^2-1=0[/TEX]
Đặt [TEX]2y=x^2 \geq 0 \Rightarrow 8y^3-6y=1 \Leftrightarrow 4y^3-3y=\frac{1}{2}[/TEX]
Dễ thấy y \geq 1 \Rightarrow PTVN \Rightarrow 0 \leq y \leq 1
Đặt [TEX]y=cost,t \in [-\frac{\pi}{2};\frac{\pi}{2}] \Rightarrow 4cos^3t-3cost=\frac{1}{2}[/TEX]
[TEX]\Leftrightarrow cos3t=\frac{1}{2} \Leftrightarrow \left[\begin{t=\frac{\pi}{9}+\frac{k2\pi}{3}}\\{t=\frac{-\pi}{9}+\frac{\k2\pi}{3}}[/TEX]
[TEX]t \in [-\frac{\pi}{2};\frac{\pi}{2}] \Rightarrow t=\frac{\pi}{9},t=-\frac{\pi}{9},t=\frac{5\pi}{9}[/TEX]
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