Question

решите пожалуйста 54 (1,6) 55(3)

Answer 1

1) lg((3x²+28)/(3x-2)) = 1

(3x²+28)/(3x-2) = 10

3x² + 28 - 30x + 20 = 0

x² - 10x + 16 = 0

x1 = 2; x2 = 8

6) lg((x²+1)/(x-2)) = 1

(x²+1)/(x-2) = 10

x² + 1 - 10x + 20 = 0

x² - 10x + 21 = 0

x1 = 3; x2 = 7

55.3) log_2(x) = a

√(1+a) + √(2a-2) = 4

1 + a + 2a - 2 + 2√(2(a²-1)) = 16

2√(2(a²-1)) = 17 - 3a > 0 => a < 17/3

8a² - 8 = 289 + 9a² - 102a

a² - 102a + 297 = 0

a = 3 (второй корень не в ОДЗ)

log_2(x) = 3

x = 2³ = 8

Answer 2

$54.1)\; \; lg(3x^2+28)-lg(3x-2)=1\; ,\; \; ODZ:\; \left \{ {{3x^2+28>0} \atop {3x-2>0}} \right. \; \Rightarrow \; x>\frac{2}{3}\\\\lg\frac{3x^2+28}{3x-2}=lg10\; \; \Rightarrow \; \; \frac{3x^2+28}{3x-2}=10\; ,\; \; \frac{3x^2+28-30x+20}{3x-2} =0\; ,\\\\3x^2-30x+48=0\; ,\; \; x^2-10x+16=0\; ,\; \; \underline {x_1=2\; ,\; x_2=8}\\\\\\54.6)\; \; lg(x^2+1)-lg(x-2)=1\; ,\; \; ODZ:\; \left \{ {{x^2+1>0} \atop {x-2>0}} \right. \; \Rightarrow \; x>2$

$lg\frac{x^2+1}{x-2}=lg10\; ,\; \; \frac{x^2+1}{x-2}=10\; ,\; \frac{x^2+1-10x+20}{x-2} =0\; ,\; \frac{x^2-10x+21}{x-2}=0\; ,\\\\x^2-10x+21=0\; \; \Rightarrow \; \; \; \underline {x_1=3\; ,\; \; x_2=7}\; \; (teor.\; Vieta)\\\\\\55.3)\; \; 2^{lg(x^2-6x+10\sqrt{10})}=2\sqrt2\; ,\; \; ODZ:\; x^2-6x+10\sqrt{10}>0\; ,\\\\D<0\; \to \; \; x\in R\\\\2^{lg(x^2-6x+10\sqrt{10})}=2^{3/2}\; ,\; \; lg(x^2-6x+10\sqrt{10})=\frac{3}{2}\; ,\\\\x^2-6x+10\sqrt{10}=10^{3/2}\; ,\; \; x^2-6x+10\sqrt{10}=10\sqrt{10}\; ,\\\\x^2-6x=0\; ,\; \; x(x-6)=0\; ,\\\\\underline {x_1=0\; ,\; \; x_2=6}$