Question

Ajutati ma sa rezolv exercitiile :D  Sunt ex de clasa a 10-a
1)2^x+1 <= 4
2)2^-x < 2^x
3)Radical din 5-x= 3x-1
4) 9^x-2 = 3 ^2-x
5) 64^x=1 pe 8
6)Radical din x+2=x
7) 125^x <= 1 pe 5
8) 2^x < 2 ^ 4 pe x

Answer 1

[tex]1)2^{x+1}\leq 4\\ 2^{x+1}\leq 2^2\\ x+1\leq 2\\ x\leq 1\Rightarrow x\in (-\infty,1]\\ \\ 2)2^{-x}\ \textless \ 2^x\\ \frac{1}{2^x}\ \textless \ 2^x\\ (2^x)^2-1\ \textgreater \ 0\\ (2^x-1)(2^x+1)\ \textgreater \ 0\\ 2^x+1\ \textgreater \ 0\forall x\in \mathbb{R}\Rightarrow 2^x-1\ \textgreater \ 0\\ 2^x\ \textgreater \ 1\\ 2^x\ \textgreater \ 2^0\Rightarrow x\ \textgreater \ 0\Rightarrow x\in (0,\infty)\\ \\ 3)\sqrt{5-x}=3x-1\\ C.E.:5-x\geq 0\Rightarrow x\leq 5\\ 5-x=(3x-1)^2\\ 5-x=9x^2-6x+1\\ 9x^2-5x-4=0\\ 9x^2-9x+4x-4=0\\ 9x(x-1)+4(x-1)=0\\ (x-1)(9x+4)=0\Rightarrow x_1=1, x_2=-\frac{4}{9}\\ [/tex]
[tex]4)9^{x-2}=3^{2-x}\\ 3^{2(x-2)}=3^{2-x}\\ 2(x-2)=2-x\\ 2x-4=2-x\\ 3x=6\Rightarrow x=2\\ \\ 5)64^x=\frac{1}{8}\\ 8^{2x}=8^{-1}\\ 2x=-1\Rightarrow x=-\frac{1}{2}\\ \\ 6)\sqrt{x+2}=x\\ C.E.: x+2\geq 0\Rightarrow x\geq -2\\ x+2=x^2\\ x^2-x-2=0\\ x^2-2x+x-2=0\\ x(x-2)+(x-2)=0\\ (x-2)(x+1)=0\Rightarrow x_1=2,x_2=-1\\ [/tex]
[tex]7)125^x\leq\frac{1}{5}\\ 5^{3x}\leq 5^{-1}\\ 3x\leq -1\\ x\leq -\frac{1}{3}\Rightarrow x\in(-\infty,-\frac{1}{3}]\\\ \\ 8)2^x\ \textless \ 2^\frac{4}{x}\\ x\ \textless \ \frac{4}{x}\\ x^2\ \textless \ 4\\ |x|\ \textless \ 2\Rightarrow x\in (-2,2)[/tex]