Question

cine ma ajuta si pe mine va rog din tot sufletul

Answer 1

 
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) 

$\it T_5 = T_{4+1} =C^4_{10} 1^6\cdot x^4 =42 x^4$

7)

[tex]\it a= \sqrt2 = 2^{\frac{1}{2}} \Rightarrow a^6 =2^{\frac{1}{2}\cdot6} =2^3=8 [/tex]

$\it b = \sqrt[3]3 \Rightarrow b = 3^{\frac{1}{3}} \Rightarrow b^6 =3^{\frac{1}{3}\cdot 6} =3^2 =9$


$c=\sqrt[6]6 = 6^{\frac{1}{6}} \Rightarrow c^6 = 6^{\frac{1}{6}\cdot 6} = 6$

Așadar, avem:

$\it c^6 \ \textless \ a^6 \ \textless \ b^6 \Longrightarrow c \ \textless \ a \ \textless \ b$

11)

$\it 2^{x+1} +2^{2-x} = 9 \Leftrightarrow 2\cdot2^x+2^2\cdot2^{-x} =9 \Leftrightarrow 2\cdot2^x + 4\dfrac{1}{2^x} =9$

$\it Notam \ 2^x = t,\ t>0\ iar\ ecuatia\ devine:$

$\it2t^2 -9t+4=0 \Rightarrow t_1 = \dfrac{1}{2}, \ \ \ t_2 = 4$

Revenim  asupra notației și obținem:

$\it 2^x = \dfrac{1}{2} \Rightarrow 2^x = 2^{-1} \Rightarrow x = -1$

$\it 2^x = 4 \Rightarrow 2^x =2^2\Rightarrow x = 2$

Deci, ecuația admite două soluții :

$\it x_1 = -1, \ \ \ x_2 = 2.$

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]