Question

Exercitiul unu si trei daca sa poate.Multumesc!

Answer 1

$\displaystyle 1)A=\{x \in Z/|x+1| \leq 24\} \\ \\ |x+1| \leq 24 \\ \\ x+1 \leq 24 \Rightarrow x \leq 24-1 \Rightarrow x \leq 23 \\ \\ x+1 \geq -24 \Rightarrow x \geq -24-1 \Rightarrow x \geq -25 \\ \\ -25 \leq x \leq 23~~~~~~~~~~~~~~~~~~~~~~~x \in [-25,23]$

$\displaystyle 3)log_3\left(x^2-4x+4\right)=2 ~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x^2-4x+4\ \textgreater \ 0 \\ \\ log_3(x^2-4x+4)=log_39 \\ \\ x^2-4x+4=9 \\ \\ x^2-4x+4-9=0 \\ \\ x^2-4x-5=0 \\ \\ a=1,~b=-4,~c=-5 \\ \\ \Delta=b^2-4ac=(-4)^2-4 \cdot 1 \cdot (-5)=16+20=36\ \textgreater \ 0 \\ \\ x_1= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-4)+ \sqrt{36} }{2 \cdot 1} = \frac{4+6}{2} = \frac{10}{2} =5 \\ \\ x_2= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-4)- \sqrt{36} }{2 \cdot 1} = \frac{4-6}{2} = \frac{-2}{2} =-1$

$\displaystyle x_1=5 \Rightarrow 5^2-4 \cdot 5+4\ \textgreater \ 0 ~A \Rightarrow x_1=5~este~solutie~a~ecuatiei \\ \\ x_2=-1 \Rightarrow (-1)^2-4 \cdot (-1)+4\ \textgreater \ 0~A \Rightarrow x_2=-1 ~este~solutie~a~ecuatiei \\ \\ S=\{-1;5\}$