having trouble understanding the concept to solve tehse kind

[spacewater]

problem
\(\displaystyle \frac{(x+y)^3-x^3}{h}\)

\(\displaystyle \frac{(x+y-x)(x^2-xy+y^2+x^2)}{y}\)

\(\displaystyle \frac{(y)(2x^2-xy+y^2)}{y}\)

final answer
\(\displaystyle (2x^2-xy+y^2)\)



problem 2
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

steps
\(\displaystyle \frac{\frac{1}{x^2+2xh+h^2}-\frac{2xh+h^2+1}{x^2+2xh+h^2}}{h}\)

\(\displaystyle (-)\frac{2xh-h^2}{x^2+2xh+h^2} \cdot \frac{1}{h}\)

final answer
\(\displaystyle (-)\frac{h(2x-h)}{x^2h+2xh^2+h^3}\)

Can someone who isnt too busy tell me what i did wrong along the steps?

 

[Deleted member 4993]

Re: correct answer to this problem?

problem
\(\displaystyle \frac{(x+y)^3-x^3}{h}\)

\(\displaystyle \frac{(x+y-x)(x^2-xy+y^2+x^2)}{y}\) .... you are almost correct - not quiet

a[sup:1o3hxlf8]3[/sup:1o3hxlf8] - b[sup:1o3hxlf8]3[/sup:1o3hxlf8] = (a - b)(a[sup:1o3hxlf8]2[/sup:1o3hxlf8] + ab + b[sup:1o3hxlf8]2[/sup:1o3hxlf8])

(x+y)[sup:1o3hxlf8]3[/sup:1o3hxlf8] - x[sup:1o3hxlf8]3[/sup:1o3hxlf8] = (x+y - x)[(x+y)[sup:1o3hxlf8]2[/sup:1o3hxlf8] + x(x+y) + x[sup:1o3hxlf8]2[/sup:1o3hxlf8]]



\(\displaystyle \frac{(y)(2x^2-xy+y^2)}{y}\)

final answer
\(\displaystyle (2x^2-xy+y^2)\)

 

[pka]

problem 2
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

Here is a suggestion. Rid the problem of so many fractions first.
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\frac{x^2 -(x+h)^2}{x^2 h(x+h)^2}\)

 

[spacewater]

Re: correct answer to this problem?

problem
\(\displaystyle \frac{(x+y)^3-x^3}{h}\)

\(\displaystyle \frac{(x+y-x)(x^2-xy+y^2+x^2)}{y}\) .... you are almost correct - not quiet

a[sup:3kcr3w7h]3[/sup:3kcr3w7h] - b[sup:3kcr3w7h]3[/sup:3kcr3w7h] = (a - b)(a[sup:3kcr3w7h]2[/sup:3kcr3w7h] + ab + b[sup:3kcr3w7h]2[/sup:3kcr3w7h])

(x+y)[sup:3kcr3w7h]3[/sup:3kcr3w7h] - x[sup:3kcr3w7h]3[/sup:3kcr3w7h] = (x+y - x)[(x+y)[sup:3kcr3w7h]2[/sup:3kcr3w7h] + x(x+y) + x[sup:3kcr3w7h]2[/sup:3kcr3w7h]]



\(\displaystyle \frac{(y)(2x^2-xy+y^2)}{y}\)

final answer
\(\displaystyle (2x^2-xy+y^2)\)

\(\displaystyle \frac {y(x^2+2xy+y^2+x^2+xy+x^2)}{y}\)

the final answer would be

\(\displaystyle 3x^2+3xy+y^2\)

 

[spacewater]

problem 2
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

Here is a suggestion. Rid the problem of so many fractions first.
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\frac{x^2 -(x+h)^2}{x^2 h(x+h)^2}\)

I don't understand how you got x^2 - (x+h)^2 for the numerator. Can you elaborate this to me in more detail if you are not too busy

 

[Deleted member 4993]

problem 2
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

steps
\(\displaystyle \frac{\frac{1}{x^2+2xh+h^2}-\frac{2xh+h^2+1}{x^2+2xh+h^2}}{h}\) <<<<<<incorrect

\(\displaystyle (-)\frac{2xh-h^2}{x^2+2xh+h^2} \cdot \frac{1}{h}\)

final answer
\(\displaystyle (-)\frac{h(2x-h)}{x^2h+2xh^2+h^3}\) <<<<<<incorrect
______________________________________________________________________

\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

\(\displaystyle = \frac{ (\frac {1}{x+h} + \frac{1}{x})\cdot (\frac {1}{x+h} - \frac{1}{x})}{h}\)

\(\displaystyle = \frac{(\frac {2x + h}{x\cdot (x+h)})\cdot (-\frac {h}{x\cdot (x+h)})}{h}\)

\(\displaystyle = -\frac {2x + h}{x^2\cdot (x+h)^2}\)

Can someone who isnt too busy tell me what i did wrong along the steps?

 

[pka]

problem 2
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

Here is a suggestion. Rid the problem of so many fractions first.
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\frac{x^2 -(x+h)^2}{x^2 h(x+h)^2}\)

I don't understand how you got x^2 - (x+h)^2 for the numerator. Can you elaborate this to me in more detail if you are not too busy

You do not understand how to handle complex fractions??
\(\displaystyle \frac{{\frac{a}{b} - \frac{c}{d}}}{h} = \frac{{ad - bc}}{{bdh}}\)

 

[Deleted member 4993]

problem 2
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)

Here is a suggestion. Rid the problem of so many fractions first.
\(\displaystyle \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\frac{x^2 -(x+h)^2}{x^2 h(x+h)^2}\)

I don't understand how you got x^2 - (x+h)^2 for the numerator.

Multiply numerator and the denomiator by x[sup:2afp49ea]2[/sup:2afp49ea](x+h)[sup:2afp49ea]2[/sup:2afp49ea]
--- the LCD --- this is a standard method for "getting rid" of fractions




Can you elaborate this to me in more detail if you are not too busy