Question

Здравствуйте.Помогите,пожалуйста.Задания во вложении/2/

Answer 1

$1)quad log_2^2x-4log_2x-1=0; ,; ; ; ; ODZ;; x textgreater 0\\t=log_2x; ,; ; ; t^2-4t-1=0\\D/4=4+1=5; ,; ; t_1=2-sqrt5; ,; ; t_2=2+sqrt5\\a); ; ; log_2x=2-sqrt5; ,; ; x=2^{2-sqrt5}\\b); ; ; log_2x=2+sqrt5; ,; ; x=2^{2+sqrt5}\\c); ; x_1cdot x_2=2^{2-sqrt5}cdot 2^{2+sqrt5}=2^{2-sqrt5+2+sqrt5}=2^4=16$

$2)quad (cosfrac{pi}{6})^{x^2-3x-10}=(sinfrac{pi}{6})^{x^2-3x-10}\\(frac{sqrt3}{2})^{x^2-3x-10} textless (frac{1}{2})^{x^2-3x-10}; |:(frac{1}{2})^{x^2-3x-10}\\(sqrt3)^{x^2-3x-10} textless 1\\(sqrt3})^{x^2-3x-10} textless (sqrt3})^0\\sqrt3 textgreater 1quad Rightarrow ; ; ; x^2-3x-10 textless 0; ,; ; ; ; x_1=-2; ,; ; x_2=5\\(x-5)(x+2) textless 0+++(-2)---(5)+++\\xin (-2,5)$

$3); ; ; lg(x^2+2x-3)=lg(x-3); ,\\ODZ:; ; left { {{x^2+2x-3 textgreater 0} atop {x-3 textgreater 0}} right. ; left { {{(x-1)(x+3) textgreater 0} atop {x textgreater 3}} right. ; left { {{xin (-infty ,-3)cup (1,+infty )} atop {xin (3,+infty )}} right. ; ; to \\xin (3,+infty )\\\x^2+2x-3=x-3\\x^2+x=0\\x(x+1)=0\\x_1=0notin ODZ; ,\\ x_2=-1notin ODZ\\Otvet:; ; xin varnothing ; .$