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
- Предмет:
  Алгебра
- Автор:
  boozerjjot
- 6 лет назад

Ответы 2

log(5)(2-x)+0,5log(5)(4x-11)=0{2-x>0⇒x<2{4x-11≠0⇒x≠2,75x∈(-∞;2)log(5)(2-x)+log(5)√(4x-11)²=0log(5)(2-x)+log(5)|4x-11|=0log(5)[(2-x)*|4x-11|]=0(2-x)*|4x-11|=1x∈(-∞;2)⇒|4x-11|=11-4x(2-x)(11-4x)=0x=2не удов услх=2,75 не удов услОтвет нет решенияlgx²+lg(x+4)²≥-lg1/9{x≠0{x≠-4x∈(-∞;-4) U (-4;0) U (0;∞)lg[x²(x+4)²]≥lg9x²(x+4)²≥9x²(x+4)²-9≥0(x(x+4)-3)(x(x+4)+3)≥0(x²+4x-3)(x²+4x+3)≥0x²+4x-3=0D=16+12=28x1=(-4-2√7)/2=-2-√7 U x2=-2+√7x²+4x+3=0x1+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;∞)
- Автор:
  patience
- 6 лет назад
- 0
$1)\; \; log_5(2-x)+0,5\cdot log_5(4x-11)^2=0\; ;\; \; ODZ:\; \left \{ {{x\ \textless \ 2} \atop {xe \frac{11}{4}}} ight. \; ,\; x\ \textless \ 2\\\\log_2(2-x)+log_5(4x-11)=0\; ,\; \; \; 0=log_51\\\\(2-x)(4x-11)=1\\\\8x-22-4x^2+11x=1\\\\4x^2-19x+23=0\\\\D=19^2-4\cdot 4\cdot 23=-7\ \textless \ 0\; \; \to \; \; net\; reshenij$ $2)\; \; lgx^2+lg(x+4)^2 \geq -lg\frac{1}{9}\; ,\; \; ODZ:\; \; \left \{ {{xe 0} \atop {xe -4}} ight. \\\\lg(x^2(x+4)^2) \geq lg9 \\\\x^2(x+4)^2 \geq 9\\\\\Big (x(x+4)\Big )^2-3^2 \geq 0\\\\(x^2+4x-3)(x^2+4x+3) \geq 0\\\\a)\; \; x^2+4x-3=0\; ,\\\\ D/4=4+3=7,\\\\x_1=-2-\sqrt7\approx -4,65\; ,\; \; \; x_2=-2+\sqrt7\approx 0,65\\\\b)\; \; x^2+4x+3=0\\\\x_1=-3,\; \; x_2=-1\; \; (teorema\; Vieta)\\\\c)\; \; \; (x+2+\sqrt7)(x+2-\sqrt7)(x+1)(x+3) \geq 0\\\\+++[-2-\sqrt7]--[-3\, ]+++[-1\, ]--[-2+\sqrt7\, ]++$ $xe0\; ,\; \; xe -4\\\\x\in (-\infty ;-2-\sqrt7\, ]\cup [-3;-1\, ]\cup [-2+\sqrt7;+\infty )$
- Автор:
  tiffanydean
- 6 лет назад
- 0
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