$$\frac{h}{2xh+h^2} = \frac{h}{h(2x+h)} = \frac{1}{2x+h}$$
I do not understand who the 1 comes to place can anyone give me some clarity ill jsut get lost
Same at this one $$\frac{3^3*3^{2x}+3^{2x}}{3^2*3^x-3^x} = \frac{3^{2x}(3^{3}+3^{2x}}{3^x(3^2-3^x)} = \frac{3x*7}{2}$$ but i still dont understand how it is 27+1/9-1
I understand h/h=1 and so on but x/xX+xX is 1/2x i dont understand this really... Thanks in forehand
$\frac{h}{2xh+h^2} = \frac{h}{h(2x+h)} =\frac{\not h}{\not h(2x+h)}=\frac{1}{2x+h}$
and
$\frac{3^3*3^{2x}+3^{2x}}{3^2*3^x-3^x} = \frac{3^{2x}(3^{3}+1)}{3^x(3^2-1)} = \frac{(3^x)^2(28)}{3^x(8)}=\frac{\not 3^x3^x*\not 4*7}{\not 3^x*\not 4*2}=\frac{3^x*7}{2}$