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′


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

$a + b = c$ $a + a = a c$ $c - a = b$ $b + b = b c$ $a^{2} + b = c^{3}$ $a + D = b^{2}$ $b + D = a^{2}$ $a^{'} b^{'} + D = c^{'} D$ $a b + D = c_{D}$ $a b + c = a b c$ $a^{'} b^{'} c^{'} + D = D_{a b c}$ $a b c + D = {a b c}_{D}$ $a^{2} + b^{\pm} = {D a b}^{\pm 2}$ $\sqrt{a^{b} + b^{+_{-}}} = D^{\pm}$ $\sqrt[\partial]{\frac{a}{b} + D} = \cot^{- 1} a^{'} b^{'}$
Commented by prakash jain last updated on 16/Jul/16

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