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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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