A=x+3 under root 3 ,B …
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Naveen Sharma 8 years, 6 months ago
Ans. A = {tex}x+3\sqrt 3{/tex}
B = {tex}1\over A^2{/tex} = {tex}1\over {\left( {x+3 \sqrt 3}\right )^2}{/tex}
Now {tex}A^2 = ({x+3 \sqrt 3})^2 \ ............... (1){/tex}
{tex}B^2 = ({1\over( x+3\sqrt 3)^2})^2 = {1\over (x+3\sqrt3)^4} \ ...... (2){/tex}
Adding (1) and (2), we get
=> {tex}A^2 + B^2 = {(x+3\sqrt 3)^2} + {1\over (x+3\sqrt 3)^4} {/tex}
=> {tex}{(x+3\sqrt 3)^2(x+3\sqrt 3)^4 + 1 \over (x+3\sqrt 3)^4}{/tex}
{tex}=> {(x+3\sqrt 3)^6 +1 \over (x+3\sqrt 3)^4}{/tex}
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