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Question-158961




Question Number 158961 by ajfour last updated on 11/Nov/21
Commented by ajfour last updated on 11/Nov/21
                Find r given, c.
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Find}\:{r}\:{given},\:{c}. \\ $$
Commented by 1549442205PVT last updated on 11/Nov/21
your conditon isn′t suitable .So if that  r^2 +x^2 =r^2 −1+1=r^2 ⇒x=0?
$${your}\:{conditon}\:{isn}'{t}\:{suitable}\:.{So}\:{if}\:{that} \\ $$$${r}^{\mathrm{2}} +{x}^{\mathrm{2}} ={r}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}={r}^{\mathrm{2}} \Rightarrow{x}=\mathrm{0}? \\ $$
Commented by 1549442205PVT last updated on 13/Nov/21
Ecxcuse me,i mistake!Thank you sir  for interesting problems.
$${Ecxcuse}\:{me},{i}\:{mistake}!{Thank}\:{you}\:{sir} \\ $$$${for}\:{interesting}\:{problems}. \\ $$
Answered by ajfour last updated on 12/Nov/21
(r^2 −1)(√(r^2 −1))+(√(r^2 −1))                 =r^2 (√((r+c)^2 −r^4 ))  ⇒  r^4 +r^2 =1+(r+c)^2   ⇒  ((r^6 −1)/(r^2 −1))=2+(r+c)^2   r^6 −2r^2 +1=(r^2 −1)(r+c)^2   (r+c)^2 −2(r^2 −1)                  =1+(r^2 −1)(r+c)^2   ⇒  2+(1/((r+c)^2 ))=((2r^2 )/((r+c)^2 ))+r^2   ......
$$\left({r}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{r}^{\mathrm{2}} −\mathrm{1}}+\sqrt{{r}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={r}^{\mathrm{2}} \sqrt{\left({r}+{c}\right)^{\mathrm{2}} −{r}^{\mathrm{4}} } \\ $$$$\Rightarrow\:\:{r}^{\mathrm{4}} +{r}^{\mathrm{2}} =\mathrm{1}+\left({r}+{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{{r}^{\mathrm{6}} −\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}+\left({r}+{c}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{6}} −\mathrm{2}{r}^{\mathrm{2}} +\mathrm{1}=\left({r}^{\mathrm{2}} −\mathrm{1}\right)\left({r}+{c}\right)^{\mathrm{2}} \\ $$$$\left({r}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({r}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\left({r}^{\mathrm{2}} −\mathrm{1}\right)\left({r}+{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}+\frac{\mathrm{1}}{\left({r}+{c}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{r}^{\mathrm{2}} }{\left({r}+{c}\right)^{\mathrm{2}} }+{r}^{\mathrm{2}} \\ $$$$…… \\ $$$$ \\ $$

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