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Question Number 173609 by infinityaction last updated on 14/Jul/22

Commented by Tawa11 last updated on 15/Jul/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Commented by universe last updated on 09/Aug/24

16

$$\mathrm{16} \\ $$

Answered by som(math1967) last updated on 14/Jul/22

x=((8(2−^8 (√(240))))/((16+(√(240)))(16−(√(240))))) x=((2−^8 (√(240)))/2) 1−x=^8 ((√(240))/2) (1−x)^8 =((240)/(256)) (1/(1−(1−x)^8 ))=(1/(1−((240)/(256))))=((256)/(16))=16

$${x}=\frac{\mathrm{8}\left(\mathrm{2}−^{\mathrm{8}} \sqrt{\mathrm{240}}\right)}{\left(\mathrm{16}+\sqrt{\mathrm{240}}\right)\left(\mathrm{16}−\sqrt{\mathrm{240}}\right)} \\ $$$${x}=\frac{\mathrm{2}−^{\mathrm{8}} \sqrt{\mathrm{240}}}{\mathrm{2}} \\ $$$$\mathrm{1}−{x}=^{\mathrm{8}} \frac{\sqrt{\mathrm{240}}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{8}} =\frac{\mathrm{240}}{\mathrm{256}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{\mathrm{8}} }=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{240}}{\mathrm{256}}}=\frac{\mathrm{256}}{\mathrm{16}}=\mathrm{16} \\ $$

Answered by behi834171 last updated on 14/Jul/22

(2+((240))^(1/8) )(2−((240))^(1/8) )=4−((240))^(1/4) (4−((240))^(1/4) )(4+((240))^(1/4) )=16−(√(240)) (16+(√(240)))(16−(√(240)))=256−240=16 ⇒x=((8(2−((240))^(1/8) ))/(16))=1−(((240))^(1/8) /2) (1−x)^8 =((((240))^(1/8) /2))^8 =((240)/(256))=((15)/(16)) 1−(1−x)^8 =1−((15)/(16))=(1/(16)) ⇒(1/(1−(1−x)^8 ))=16 .■

$$\left(\mathrm{2}+\sqrt[{\mathrm{8}}]{\mathrm{240}}\right)\left(\mathrm{2}−\sqrt[{\mathrm{8}}]{\mathrm{240}}\right)=\mathrm{4}−\sqrt[{\mathrm{4}}]{\mathrm{240}} \\ $$$$\left(\mathrm{4}−\sqrt[{\mathrm{4}}]{\mathrm{240}}\right)\left(\mathrm{4}+\sqrt[{\mathrm{4}}]{\mathrm{240}}\right)=\mathrm{16}−\sqrt{\mathrm{240}} \\ $$$$\left(\mathrm{16}+\sqrt{\mathrm{240}}\right)\left(\mathrm{16}−\sqrt{\mathrm{240}}\right)=\mathrm{256}−\mathrm{240}=\mathrm{16} \\ $$$$\Rightarrow\boldsymbol{{x}}=\frac{\mathrm{8}\left(\mathrm{2}−\sqrt[{\mathrm{8}}]{\mathrm{240}}\right)}{\mathrm{16}}=\mathrm{1}−\frac{\sqrt[{\mathrm{8}}]{\mathrm{240}}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\boldsymbol{{x}}\right)^{\mathrm{8}} =\left(\frac{\sqrt[{\mathrm{8}}]{\mathrm{240}}}{\mathrm{2}}\right)^{\mathrm{8}} =\frac{\mathrm{240}}{\mathrm{256}}=\frac{\mathrm{15}}{\mathrm{16}} \\ $$$$\mathrm{1}−\left(\mathrm{1}−\boldsymbol{{x}}\right)^{\mathrm{8}} =\mathrm{1}−\frac{\mathrm{15}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{1}−\boldsymbol{{x}}\right)^{\mathrm{8}} }=\mathrm{16}\:\:\:\:\:.\blacksquare \\ $$

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