Ta có \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\Rightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\Rightarrow255-2x=0\Rightarrow x=\dfrac{255}{2}\)
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