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Question Number 135378 by Bird last updated on 12/Mar/21

compare without calculator 5((√(1+(√7)))−1) and 7((√(1+(√5)))−1)

$${compare}\:{without}\:{calculator} \\ $$$$\mathrm{5}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{7}}}−\mathrm{1}\right)\:{and}\:\mathrm{7}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}−\mathrm{1}\right) \\ $$

Answered by mr W last updated on 12/Mar/21

5((√(1+(√7)))−1)<5((√(1+(√9)))−1)=5 7((√(1+(√5)))−1)>7((√(1+(√4)))−1)=7((√3)−1) =7(√3)−7=7(√((147)/(49)))−7>7(√((144)/(49)))−7 =7×((12)/7)−7=12−7=5 since 7((√(1+(√5)))−1)>5 and 5((√(1+(√7)))−1)<5, ⇒7((√(1+(√5)))−1)>5((√(1+(√7)))−1)

$$\mathrm{5}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{7}}}−\mathrm{1}\right)<\mathrm{5}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{9}}}−\mathrm{1}\right)=\mathrm{5} \\ $$$$ \\ $$$$\mathrm{7}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}−\mathrm{1}\right)>\mathrm{7}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{4}}}−\mathrm{1}\right)=\mathrm{7}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$=\mathrm{7}\sqrt{\mathrm{3}}−\mathrm{7}=\mathrm{7}\sqrt{\frac{\mathrm{147}}{\mathrm{49}}}−\mathrm{7}>\mathrm{7}\sqrt{\frac{\mathrm{144}}{\mathrm{49}}}−\mathrm{7} \\ $$$$=\mathrm{7}×\frac{\mathrm{12}}{\mathrm{7}}−\mathrm{7}=\mathrm{12}−\mathrm{7}=\mathrm{5} \\ $$$$ \\ $$$${since}\:\mathrm{7}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}−\mathrm{1}\right)>\mathrm{5}\:{and} \\ $$$$\mathrm{5}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{7}}}−\mathrm{1}\right)<\mathrm{5}, \\ $$$$\Rightarrow\mathrm{7}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}−\mathrm{1}\right)>\mathrm{5}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{7}}}−\mathrm{1}\right) \\ $$

Commented by mathmax by abdo last updated on 12/Mar/21

thankx sir mrw

$$\mathrm{thankx}\:\mathrm{sir}\:\mathrm{mrw} \\ $$

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