Question

Решите уравнение и неравенство пожалуйста

Answer 1

$sqrt{x^2-2x-4}=sqrt{2x^2-6x-1}, \ left { {{x^2-2x-4geq0,} atop {2x^2-6x-1geq0;}} right. left { {{D=5, left [ {{xleq1-sqrt{5},} atop {xgeq1+sqrt{5};}} right. } atop {D=11, left [ {{xleqfrac{3-sqrt{11}}{2},} atop {xgeqfrac{3+sqrt{11}}{2}}} right. }} right. left [ {{xleq1-sqrt{5},} atop {xgeq1+sqrt{5};}} right. \ x^2-2x-4=2x^2-6x-1, \ x^2-4x+3=0, \ x_1=1, x_2=3, \ 1-sqrt{5} textless 1 textless 1+sqrt{5}, \ 1-sqrt{5} textless 3 textless 1+sqrt{5}, \ xinvarnothing.$


$log_xfrac{2x+5}{4(x-1)}leq0, \ left { {{frac{2x+5}{4(x-1)} textgreater 0,} atop {4(x-1) neq 0;}} right. \ (2x+5)(x-1) textgreater 0, \ left [ {{x textless -2,5,} atop {x textgreater 1;}} right.$
$left [ {{ left { {{0 textless x textless 1,} atop {frac{2x+5}{4(x-1)} geq 1,}} right. } atop {left { {{x textgreater 1,} atop {frac{2x+5}{4(x-1)}leq1,}} right.}} right. left [ {{ left { {{0 textless x textless 1,} atop {frac{2x+5}{4(x-1)}-1geq0,}} right. } atop {left { {{x textgreater 1,} atop {frac{2x+5}{4(x-1)}-1leq0,}} right.}} right. left [ {{ left { {{0 textless x textless 1,} atop {frac{-2x+9}{4(x-1)}geq0,}} right. } atop {left { {{x textgreater 1,} atop {frac{-2x+9}{4(x-1)}leq0,}} right.}} right.$
$left [ {{ left { {{0 textless x textless 1,} atop {(2x-9)(x-1)leq0,}} right. } atop {left { {{x textgreater 1,} atop {(2x-9)(x-1)geq0,}} right.}} right. left [ {{ left { {{0 textless x textless 1,} atop {1 textless xleq 4,5,}} right. } atop {left { {{x textgreater 1,} atop { left [ {{x textless 1,} atop {x geq 4,5;}} right. }} right.}} right. left [ {{xinvarnothing,} atop {x geq 4,5; }} right.}} right. \ x geq 4,5.$