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Question-156754




Question Number 156754 by abdurehime last updated on 15/Oct/21
Answered by MathsFan last updated on 15/Oct/21
(√(5+2(√6)))=(√(5+(√(24))))=(√((5+(√1))/2))+(√((5−(√1))/2))=(√3)+(√2)    (√(5−2(√6)))=(√((5+(√1))/2))−(√((5−(√1))/2))=(√3)−(√2)   i. (√3)+(√2)−(√3)+(√2)=  2(√2)    ii.   { ((. (√(7−(√(24))))=(√((7+5)/2))−(√((7−5)/2))=(√6)−1)),(( (√(7+(√(24))))= (√6)+1)) :}    ∴ (((√6)+1+(√6)−1)/2) = ((2(√6))/2) = (√6)   stay safe     stay blessed...
$$\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}=\sqrt{\mathrm{5}+\sqrt{\mathrm{24}}}=\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{1}}}{\mathrm{2}}}+\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{1}}}{\mathrm{2}}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$$\:\:\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}=\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{1}}}{\mathrm{2}}}−\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{1}}}{\mathrm{2}}}=\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$$\:{i}.\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}=\:\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{{ii}}. \\ $$$$\begin{cases}{.\:\sqrt{\mathrm{7}−\sqrt{\mathrm{24}}}=\sqrt{\frac{\mathrm{7}+\mathrm{5}}{\mathrm{2}}}−\sqrt{\frac{\mathrm{7}−\mathrm{5}}{\mathrm{2}}}=\sqrt{\mathrm{6}}−\mathrm{1}}\\{\:\sqrt{\mathrm{7}+\sqrt{\mathrm{24}}}=\:\sqrt{\mathrm{6}}+\mathrm{1}}\end{cases} \\ $$$$\:\:\therefore\:\frac{\sqrt{\mathrm{6}}+\mathrm{1}+\sqrt{\mathrm{6}}−\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}}\:=\:\sqrt{\mathrm{6}} \\ $$$$\:\boldsymbol{\mathrm{stay}}\:\boldsymbol{\mathrm{safe}}\:\: \\ $$$$\:\boldsymbol{\mathrm{stay}}\:\boldsymbol{\mathrm{blessed}}… \\ $$$$\: \\ $$
Commented by abdurehime last updated on 15/Oct/21
please help question 4
$$\mathrm{please}\:\mathrm{help}\:\mathrm{question}\:\mathrm{4} \\ $$

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