Punctele c) si d) ... multumesc!

c)(4x-5-3)(4x-5+3)/(4x-3-5)(4x-3+5)=
(4x-8)(4x-2)/(4x-8)(4x+2)=
(4x-2)/(4x+2)=
2(2x-1)/2(2x+1)=
(2x-1)/(2x+1)
Prima data am simplificat cu (4x-8), apoi cu 2, deci 2(4x-8)=8x-16

d) 2^x•(1+2+2^2)=2•7
2^x•7=2•7 |:7
2^x=2
=> x=1

   
[tex]\displaystyle\\ c)\\\\ \frac{(4x-5)^2-9}{(4x-3)^2-25}=\\\\ =\frac{(4x-5)^2-3^2}{(4x-3)^2-5^2}=\\\\ =\frac{(4x-5-3)(4x-5+3)}{(4x-3-5)(4x-3+5)}=\\\\ =\frac{(4x-8)(4x-2)}{(4x-8)(4x+2)}=\\\\ =\frac{(4x-8)\cdot 2(2x-1)}{(4x-8)\cdot 2(2x+1)}=\\\\ =\frac{[(4x-8)\cdot 2](2x-1)}{[(4x-8)\cdot 2](2x+1)}=\\\\ =\frac{(8x-16)(2x-1)}{(8x-16)(2x+1)}=~\text{ Simplificam cu }8x-16. \\\\ =\boxed{\frac{2x-1}{2x+1}} [/tex]


[tex]\displaystyle\\ c)\\ 2^x+2^{x+1}+2^{x+2}=14\\\\ 2^x+2^x\cdot 2^1+2^x\cdot 2^2=14\\\\ 2^x(1 + 2^1 + 2^2)=14\\\\ 2^x(1+2+4)= 14\\\\ 2^x \cdot 7 = 14\\\\ 2^x = \frac{14}{7} \\\\ 2^x = 2\\\\ 2^x = 2^1\\\\ \boxed{x=1} [/tex]