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If $2, 1, 1$ are roots of the equation $x^3-4x^2+5x-2=0$,then find the roots of the equation $\left(x+\frac{1}{3}\right)^3-4\left(x+\frac{1}{3}\right)^2+5\left(x+\frac{1}{3}\right)-2=0$.

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The equation $\sqrt{3x^2 + x + 5} = x - 3$,where $x$ is real,has

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