Question

Решите неравенство
$( {x}^{2} - 8x + 7)* sqrt{ log_{5}( {x}^{2} - 3 ) } leqslant 0$

Answer 1

$(x^2 - 8x + 7) cdot sqrt{ log_{5}(x^2 - 3 )} le 0$

Область определения $sqrt{ log_{5}(x^2 - 3 )}$

$begin{cases}x^2 - 3>0\ log_{5}(x^2 - 3 ) ge 0end{cases}$

$begin{cases}(x- sqrt{3} )(x+ sqrt{3} )>0\ log_{5}(x^2 - 3 ) ge log_51end{cases}$

$begin{cases}x in (- infty ;- sqrt{3} ) cup ( sqrt{3};+ infty )\ x^2 - 3 ge1end{cases}$

$begin{cases}x in (- infty ;- sqrt{3} ) cup ( sqrt{3};+ infty )\ x^2 - 4 ge0end{cases}$

$begin{cases}x in (- infty ;- sqrt{3} ) cup ( sqrt{3};+ infty )\(x-2)(x+2) ge0end{cases}$

$begin{cases}x in (- infty ;- sqrt{3} ) cup ( sqrt{3};+ infty )\x in left(- infty ;-2 right] cup left[2;+ inftyright) end{cases}$

$x in left(- infty ;-2 right] cup left[2;+ inftyright)$

$x in left(- infty ;-2 right] cup left[2;+ inftyright) Rightarrow sqrt{ log_{5}(x^2 - 3 )} ge 0$

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$x^2 - 8x + 7 le 0$

$D=(-8)^2-4 cdot 1 cdot 7=64-28=36$

$sqrt{D}= sqrt{36} =6$

$x_1= frac{8-6}{2} = frac{2}{2}=1$

$x_2= frac{8+6}{2} = frac{14}{2}=7$

$x in left[ 1;7right]$

$begin{cases} x in left(- infty ;-2 right] cup left[2;+ inftyright)\ x in left[ 1;7right]end{cases}$

$x in left[ 2;7right]$

Ответ

$x inleft{-2 right} cup left[ 2;7right]$