Question

Объясните подробно как решить это неравенство (1-x^2)*√(2x-1) ≤0

Answer 1

$(1- x^{2} ) \sqrt{2x-1} \leq 0 \\ OD3:2x-1 \geq 0 \\ x \geq \frac{1}{2} \\ (1- x^{2} ) \sqrt{2x-1} =0 \\ 1- x^{2} =0 =\ \textgreater \ x_{1,2} =+-1 \\ \sqrt{2x-1} =0 =\ \textgreater \ x_{3} = \frac{1}{2}$
-1 не входит в ОДЗ
  .________.______
1/2        -      1    +
ОТВЕТ: {1/2}; [1; +∞)

Answer 2

$(1- x^{2} )* \sqrt{2x-1} \leq 0 \\ \left \{ {{1- x^{2} \leq 0} \atop {2x-1 \geq 0}} \right. \\ \left \{ {{x^{2} - 1 \geq 0} \atop {x \geq \frac{1}{2} }} \right. \\$
$\left \{ {{(x- 1)(x+1) \geq 0} \atop {x \geq \frac{1}{2} }} \right. \\ \left \{ {{x \leq -1 U x \geq 1} \atop {x \geq \frac{1}{2} }} \right. \\ x \geq 1$

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ответ:  [  1 ;  + оо  )