Lg(x-6) -(1/2)*Lg2 ≥Lg3 +(1/2)*Lg(x-10) ;
* * * ОДЗ : { x-6 >0 ; x-10 >0 .⇒{x>6 ; x>10 .⇒ x >10.
x∈ (10 ; ∞).
-----
Lg(x-6) ≥Lg3 +(1/2)*Lg(x-10) +(1/2)Lg2 ;
Lg(x-6) ≥Lg3+ Lg√(x-10) +Lg√2 ;
Lg(x-6) ≥Lg( 3*√(x-10) *√2 ) ; * * * т.к. основание логарифма =10 >1,то
x -6 ≥ 3√2* √(x-10) ; т.к. x >10 то
(x -6)² ≥ (3√2* √(x-10) ) ²
x² -12x+36 ≥18(x -10) ;
x² -30x +216 ≥ 0 ;
(x-12)(x-18) ≥ 0 ;
методом интервалов :
+ - +
(10) ------ [12] ---- [18] -----
ответ: x ∈ (10 ;12] U [18 ;∞) .
* * *
Lg(x-6) -(1/2)*Lg2 ≥Lg3 +(1/2)*Lg(x-10). ⇔
{ x-6 >0 ; x-10 >0 ; x-6 ≥ 3√2*√(x-10).⇔ { x > 10 ; (x-6)² ≥ 3√2*(x-10).
Answers & Comments
Verified answer
ОДЗ:
а) x-6>0
x>6
b) x-10>0
x>10
В итоге ОДЗ: x>10
{x-6≥0
{18(x-10)≤(x-6)²
{18(x-10)≥0
a) x-6≥0
x≥6
b) 18(x-10)≤(x-6)²
18x-180≤x²-12x+36
-x²+18x+12x-180-36≤0
-x²+30x-216≤0
x²-30x+216≥0
D=900-4*216=36
x₁=(30-6)/2=12
x₂=(30+6)/2=18
+ - +
-------- 12 ------------- 18 -----------
\\\\\\\\\ \\\\\\\\\\\\
x∈(-∞; 12]U[18; +∞)
c) 18(x-10)≥0
x-10≥0
x≥10
Объединяем ОДЗ и решения трех неравенств в систему:
{x>10
{x≥6
{x∈(-∞; 12]U[18; +∞)
{x≥10
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
-------- 10 ------ 12 -------------- 18 ------------
/////////////////////// ////////////////////////
x∈(10; 12]U[18; +∞)
Ответ: (10; 12]U[18; +∞)