Question

ПОЖАЛУЙСТА СРОЧНО НАДО!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!50 БАЛОВ

Answer 1

Ответ:

A) x = 0; 2

Б) x = 0; 1

Объяснение:

$\textbf{a)}\ 2^x+2^{2-x}=5\\2^x+\frac{2^2}{2^x} =5\\2^x=t > 0\\t+\frac{4}{t} =5\quad |\times t\\t^2+4=5t\\t^2-5t+4=0\\t_{1,2}=(1;4)\\\\\textbf{I)}\ t=1\\2^x=1\\\fbox{x=0}\\\\\textbf{II)}\ t=4\\2^x=4=2^2\\\fbox{x=2}$

$\textbf{B)}\ 4*9^x-7*12^x+3*16^x=0\\4*(3^x)^2-7*(3*4)^x+3*(4^x)^2=0\quad |\div(4^x)^2\neq 0\\4*((\frac{3}{4})^x) ^2-7*(\frac{3}{4})^x +3=0\\(\frac{3}{4})^x =t > 0\\4t^2-7t+3=0\\t_{1,2}=(\frac{3}{4};\ 1)\\\\\textbf{I)}\ t = \frac{3}{4} \\(\frac{3}{4} )^x=\frac{3}{4} \\\fbox{x=1}\\\\\textbf{II)}\ t = 1\\(\frac{3}{4} )^x=1\\\fbox{x=0}$