[Toán] $3x (2 + \sqrt{9x^2 + 3} ) + ( 4x + 2)(\sqrt{1 + x + x^2} + 1 ) = 0$

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tranvanhung7997


$$3x (2 + \sqrt{9x^2 + 3} ) + ( 4x + 2)(\sqrt{1 + x + x^2} + 1 ) = 0$$
$$\leftrightarrow (2x+1)(\sqrt{4x^2+4x+4}+2) = - 3x(\sqrt{9x^2+3}+2)$$
$$\leftrightarrow (2x+1)(\sqrt{(2x+1)^2+3}+2) = (-3x)(\sqrt{(-3x)^2+3}+2)$$
Xét hàm số: $f(u)= u(\sqrt{u^2+3}+2) = 2u + u\sqrt{u^2+3}$
Ta có: $f'(u) = 2+ \sqrt{u^2 +3} + \dfrac{u^2}{\sqrt{u^2+3}} > 0$
Suy ra hàm số đồng biến trên R
Do đó: $$f(2x+1)=f(-3x) \leftrightarrow 2x+1=-3x \leftrightarrow x=\dfrac{-1}{5}$$
 
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