Question

Непростой пример из профильного уровня ЕГЭ по математике

Answer 1

$log, _{|x-5|}, (2x^2-10x+8) leq 2; ,\\ODZ:; ; left { {{|x-5| textgreater 0; ,; |x-5|ne 1} atop {2x^2-10x+8 textgreater 0}} right. ; ; left { {{xne 4; ,; xne 6} atop {xin (-infty ,1)cup (4,+infty )}} right. \\1); ; 0 textless |x-5| textless 1; ; Rightarrow ; ; left { {{x-5 textless 1} atop {x-5 textgreater -1}} right. ; Rightarrow ; 4 textless x textless 6\\ left { {{4 textless x textless 6} atop {xin (-infty ,1)cup (4,+infty )}} right. ; ; Rightarrow ; ; xin (4,6)\\2x^2-10x+8 geq |x-5|^2\\|x-5|^2=(x-5)^2; ; to$

$2x^2-10x+8 geq x^2-10x+25\\x^2-17 geq 0; ,; ; (x-sqrt{17})(x+sqrt{17}) geq 0; ,; ; ; sqrt{17}approx 4,12\\xin (-infty ,-sqrt{17}; ]cup [; sqrt{17},+infty )\\ left { {{xin (4,6)} atop {xin (-infty ,sqrt{17}, ]cup [, sqrt{17},+infty )}} right. ; ; to ; ; ; xin [; sqrt{17},6)\\2); ; |x-5| textgreater 1; ; Rightarrow ; ; left [ {{x-5 textgreater 1} atop {x-5 textless -1}} right. ; ; Rightarrow ; ; xin (-infty ,4)cup (6,+infty )\\2x^2-10x+8 leq |x-5|^2$

$x^2-17 leq 0; ,; ; ; xin [; -sqrt{17},sqrt{17}; ]$

$3); ; left { {{xin (-infty ,4)cup (6,+infty )} atop {xin [, -sqrt{17},sqrt{17}, ]}} right. ; ; to ; ; xin [; -sqrt{17},4) \\4); ; xin [; -sqrt{17},4)cup [; sqrt{17},6); .$

$5); ; left { {{xin (-infty ,1)cup (4,+infty ); ,; xne 6} atop {xin [, -sqrt{17},4)cup [, sqrt{17},6)}} right. ; ; to ; ; xin [, -sqrt{17},1)cup [, sqrt{17},6)$

Answer 2

Основание не =1