[Toán 9] Giải pt: $\sqrt{x^2+1}+\sqrt{x^2-x+1}+\sqrt{2x^2-x+1}=\sqrt{3(4x^2-2x+3)}$

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vansang02121998

$\sqrt{x^2+1}+\sqrt{x^2-x+1}+\sqrt{2x^2-x+1}=\sqrt{3(4x^2-2x+3)}$

$\sqrt{x^2+1}=a;\sqrt{x^2-x+1}=b;\sqrt{2x^2-x+1}=c$

$\rightarrow a^2+b^2+c^2=4x^2-2x+3$

Ta có phương trình

$a+b+c=\sqrt{3(a^2+b^2+c^2)}$

$a^2+b^2+c^2 \ge ab+ac+bc$

$\leftrightarrow 3(a^2+b^2+c^2) \ge (a+b+c)^2$

$\leftrightarrow \sqrt{3(a^2+b^2+c^2)} \ge a+b+c$

$\rightarrow a=b=c$

$\leftrightarrow x^2+1=x^2-x+1=2x^2-x+1$

$\leftrightarrow x=0$
 
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