2)
[tex]\it f(x) =\dfrac{1-x^2}{1+x^2}
\\\;\\
\it f(0) = 1
\\\;\\
\it (fof)(0) = f(f(0)) =f(1) =\dfrac{\it1-1}{\it1+1} =\dfrac{\it0}{\it2} = \it 0[/tex]
4)
[tex]\it T_5 = T_{4+1} =C^4_{10} 1^6\cdot x^4 =42 x^4[/tex]
7)
[tex]\it a= \sqrt2 = 2^{\frac{1}{2}} \Rightarrow a^6 =2^{\frac{1}{2}\cdot6} =2^3=8
[/tex]
[tex]\it b = \sqrt[3]3 \Rightarrow b = 3^{\frac{1}{3}} \Rightarrow b^6 =3^{\frac{1}{3}\cdot 6} =3^2 =9[/tex]
[tex]c=\sqrt[6]6 = 6^{\frac{1}{6}} \Rightarrow c^6 = 6^{\frac{1}{6}\cdot 6} = 6[/tex]
Așadar, avem:
[tex]\it c^6 \ \textless \ a^6 \ \textless \ b^6 \Longrightarrow c \ \textless \ a \ \textless \ b[/tex]
11)
[tex]\it 2^{x+1} +2^{2-x} = 9 \Leftrightarrow 2\cdot2^x+2^2\cdot2^{-x} =9 \Leftrightarrow 2\cdot2^x + 4\dfrac{1}{2^x} =9[/tex]
[tex] \it Notam \ 2^x = t,\ t>0\ iar\ ecuatia\ devine:[/tex]
[tex]\it2t^2 -9t+4=0 \Rightarrow t_1 = \dfrac{1}{2}, \ \ \ t_2 = 4[/tex]
Revenim asupra notației și obținem:
[tex]\it 2^x = \dfrac{1}{2} \Rightarrow 2^x = 2^{-1} \Rightarrow x = -1[/tex]
[tex]\it 2^x = 4 \Rightarrow 2^x =2^2\Rightarrow x = 2[/tex]
Deci, ecuația admite două soluții :
[tex]\it x_1 = -1, \ \ \ x_2 = 2.[/tex]
17)
[tex]\it \cos 70^0+\cos 110^0 = \cos70^0 + \cos(180^0-70^0) =
\\\;\\
\it \cos\it70^0+ \it\cos180^0\cos 70^0 +\sin180^0\sin70^0 = \cos70^0-\cos70^0=0[/tex]
