Алгебра: Log5(2-x)+0,5log5(4x-11)^2=0 lgx^2+lg(x+4)^2 =-lg1/9

Log5(2-x)+0,5log5(4x-11)^2=0 lgx^2+lg(x+4)^2> =-lg1/9

Ответ:

шынар16

05.10.2020 12:29

Log(5)(2-x)+0,5log(5)(4x-11)=0
{2-x>0⇒x<2
{4x-11≠0⇒x≠2,75
x∈(-∞;2)
log(5)(2-x)+log(5)√(4x-11)²=0
log(5)(2-x)+log(5)|4x-11|=0
log(5)[(2-x)*|4x-11|]=0
(2-x)*|4x-11|=1
x∈(-∞;2)⇒|4x-11|=11-4x
(2-x)(11-4x)=0
x=2не удов усл
х=2,75 не удов усл
ответ нет решения

lgx²+lg(x+4)²≥-lg1/9
{x≠0
{x≠-4
x∈(-∞;-4) U (-4;0) U (0;∞)
lg[x²(x+4)²]≥lg9
x²(x+4)²≥9
x²(x+4)²-9≥0
(x(x+4)-3)(x(x+4)+3)≥0
(x²+4x-3)(x²+4x+3)≥0
x²+4x-3=0
D=16+12=28
x1=(-4-2√7)/2=-2-√7 U x2=-2+√7
x²+4x+3=0
x1+x2=-4 U x1*x2=3⇒x1=-3 U x2=-1
+ _ + _ +
[-2-√7](-4)[-3][-1](0)[-2+√7]

x∈(-∞;-2-√7] U [-3;-1] U [-2+√7;∞)

0,0(0 оценок)

Ответ:

winterwhite123

05.10.2020 12:29

$1)\; \; log_5(2-x)+0,5\cdot log_5(4x-11)^2=0\; ;\; \; ODZ:\; \left \{ {{x\ \textless \ 2} \atop {x\ne \frac{11}{4}}} \right. \; ,\; x\ \textless \ 2\\\\log_2(2-x)+log_5(4x-11)=0\; ,\; \; \; 0=log_51\\\\(2-x)(4x-11)=1\\\\8x-22-4x^2+11x=1\\\\4x^2-19x+23=0\\\\D=19^2-4\cdot 4\cdot 23=-7\ \textless \ 0\; \; \to \; \; net\; reshenij$

$2)\; \; lgx^2+lg(x+4)^2 \geq -lg\frac{1}{9}\; ,\; \; ODZ:\; \; \left \{ {{x\ne 0} \atop {x\ne -4}} \right. \\\\lg(x^2(x+4)^2) \geq lg9 \\\\x^2(x+4)^2 \geq 9\\\\\Big (x(x+4)\Big )^2-3^2 \geq 0\\\\(x^2+4x-3)(x^2+4x+3) \geq 0\\\\a)\; \; x^2+4x-3=0\; ,\\\\ D/4=4+3=7,\\\\x_1=-2-\sqrt7\approx -4,65\; ,\; \; \; x_2=-2+\sqrt7\approx 0,65\\\\b)\; \; x^2+4x+3=0\\\\x_1=-3,\; \; x_2=-1\; \; (teorema\; Vieta)\\\\c)\; \; \; (x+2+\sqrt7)(x+2-\sqrt7)(x+1)(x+3) \geq 0\\\\+++[-2-\sqrt7]--[-3\, ]+++[-1\, ]--[-2+\sqrt7\, ]++$

$x\ne0\; ,\; \; x\ne -4\\\\x\in (-\infty ;-2-\sqrt7\, ]\cup [-3;-1\, ]\cup [-2+\sqrt7;+\infty )$