Question

√x-2√x-1+√x+3-4√x-1=1 корень под корьнем

Answer 1

$\sqrt{ x-2\sqrt{x-1} }+\sqrt{ x+3-4\sqrt{x-1} } =1\;,\; \; ODZ:\; \; x \geq 1\\\\ \sqrt{(1-\sqrt{x-1})^2} + \sqrt{(2-\sqrt{x-1})^2} =1\\\\|1-\sqrt{x-1}|+|2-\sqrt{x-1}|=1\\\\Zamena:\; \; t= \sqrt{x-1} \; ,\; \; |1-t|+|2-t|=1\\\\-------(1)------(2)------\ \textgreater \ t\\\\a)\; \; t\ \textless \ 1:\; \; |1-t|=1-t,\; |2-t|=2-t\\\\1-t+2-t=1\; ,\; \; -2t=-2,\; \; t=1\notin (-\infty ,1)\\\\b)\; \; 1 \leq t\ \textless \ 2:\; \; |1-t|=t-1,\; |2-t|=2-t\\\\t-1+2-t=1\; ,\; \; 1=1\; verno$

$c)\; \; t \geq 2:\; \; |1-t|=t-1,\; \; |2-t|=t-2\\\\t-1+t-2=1\; ,\; 2t=4,\; \; t=2$

$1 \leq t \leq 2\\\\1 \leq \sqrt{x-1} \leq 2\\\\a)\; \; \sqrt{x-1 }\geq 1\; ,\; \; x-1 \geq 1\; ,\; \; x \geq 2\\\\b)\; \; \sqrt{x-1}\leq 2\; ,\; \; x-1 \leq 4\; ,\; \; x \leq 5\\\\2 \leq x \leq 5\\\\x\in [\, 2,\, 5]$