Ajutati-ma cu intrebarile care cu + va rooooog frumos, macar cu o intrebare, dau coroana

[tex] 1.\\a=log_216^{\frac{1}{2}}-log_3(\frac{1}{9})^{\frac{1}{2}} =\frac{1}{2} (log_216-log_3(\frac{1}{9}))=\frac{1}{2}(4+2)=\frac{6}{2}=3.\\ 2.\\ Numerele~alea~sunt~in~progresie~aritmetica~daca~:\\ 1-x=\frac{x+1+4}{2} \Leftrightarrow~2-2x=x+5~\Leftrightarrow~x=-1.\\ 3.\\ x^2*6^x-6^{2+x}\leq 0~\Leftrightarrow~x^2*6x-6^2*6^x\leq 0~\Leftrightarrow~6^x(x^2-6^2)\leq 0\\ 6^x(x-6)(x+6)\leq 0\\ Deoarece~6^x~n-are~cum~sa~fie~\leq 0~ai~de~analizat~asta:~(x-6)(x+6)\leq 0.\\ Radacinile~sunt~x_{1,2}=+6~si~-6.\\ Daca~faci~tabel,~vei~obtine~x\in~[-6,6].\\ 4.\\ Egalezi~f(x)=y.\\ \sqrt{9-x}-\sqrt{4-x}=1~|^2\\ 9-x+4-x-2\sqrt{(9-x)(4-x)}=1\\ .....................\\ 6-x=\sqrt{x^2-13x+36} ~|^2\\ 36+x^2-12x=x^2-13x+36\\ x=0.\\ f(0)=3-2=1=y\\ Punctul~este~M(0,1).\\ [/tex]