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Question Number 6725 by Danilka last updated on 16/Jul/16

a+b=c  a+a=ac  c−a=b  b+b=bc  a^2 +b=c^3   a+D=b^2   b+D=a^2   a′b′+D=c′D  ab+D=c_D   ab+c=abc  a′b′c′+D=D_(abc)   abc+D=abc_D   a^2 +b^± =Dab^(±2)   (√(a^b +b^+_−  ))=D^±   (((a/b)+D))^(1/∂) =cot^(−1) a′b′

$${a}+{b}={c} \\ $$$${a}+{a}={ac} \\ $$$${c}−{a}={b} \\ $$$${b}+{b}={bc} \\ $$$${a}^{\mathrm{2}} +{b}={c}^{\mathrm{3}} \\ $$$${a}+{D}={b}^{\mathrm{2}} \\ $$$${b}+{D}={a}^{\mathrm{2}} \\ $$$${a}'{b}'+{D}={c}'{D} \\ $$$${ab}+{D}={c}_{{D}} \\ $$$${ab}+{c}={abc} \\ $$$${a}'{b}'{c}'+{D}={D}_{{abc}} \\ $$$${abc}+{D}={abc}_{{D}} \\ $$$${a}^{\mathrm{2}} +{b}^{\pm} ={Dab}^{\pm\mathrm{2}} \\ $$$$\sqrt{{a}^{{b}} +{b}^{+_{−} } }={D}^{\pm} \\ $$$$\sqrt[{\partial}]{\frac{{a}}{{b}}+{D}}=\mathrm{cot}^{−\mathrm{1}} {a}'{b}' \\ $$

Commented by prakash jain last updated on 16/Jul/16

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