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

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Answer 3

55. 3) ?

Answer 4

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

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