tex] , если а)x

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Упростите выражение [tex] \sqrt{4x^{2} - 4x+1} [/tex] , если а)x<0,5 б)x ≥0,5

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$4x^2-4x+1=4(x^2-x)+1=4(\, (x-\frac{1}{2})^2-\frac{1}{4}\, )+1=\\\\=4(x-\frac{1}{2})^2-1+1=4(x-\frac{1}{2})^2=4(x-0,5)^2\\\\a)\; \; x\ \textless \ 0,5\; \; \to \; (x-0,5)\ \textless \ 0\\\\\sqrt{4x^2-x+1}=\sqrt{4(x-0,5)^2}=2\cdot |x-0,5|=2\cdot (0,5-x)=1-2x\\\\b)\; \; x \geq 0,5\; \; \to \; \; (x-0,5) \geq 0\\\\\sqrt{4x^2-4x+1}=\sqrt{4(x-0,5)^2}=2\cdot |x-0,5|=2(x-0,5)=2x-1$

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