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Question-222585

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Question Number 222585 by Rojarani last updated on 30/Jun/25
Commented by AntonCWX8 last updated on 01/Jul/25
Now I′ll give you guys another challenge. What if the numbers beside the (√( )) symbol represent the nth root? Say 5(√( )) = (( ))^(1/5)
$${Now}\:{I}'{ll}\:{give}\:{you}\:{guys}\:{another}\:{challenge}. \\ $$$${What}\:{if}\:{the}\:{numbers}\:{beside}\:{the}\:\sqrt{\:\:}\:{symbol}\:{represent}\:{the}\:{nth}\:{root}? \\ $$$${Say}\:\mathrm{5}\sqrt{\:\:\:}\:=\:\sqrt[{\mathrm{5}}]{\:\:\:\:} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Jul/25
((x^2 +1)/(y^2 +1))=x^2 ...i ((y^2 +1)/(z^2 +1))=6^5 y^5 ...ii ((z^2 +1)/(x^2 +1))=20^7 z^7 i×ii×iii: 6^5 .20^7 .x^2 y^5 z^7 =1 i⇒x^2 +1=x^2 y^2 +x^2 x^2 y^2 =1 6^5 .20^7 .(x^2 y^2 )y^3 z^7 =1 y^3 z^7 =(1/(6^5 .20^7 )) ii⇒y^2 +1=6^5 y^5 z^2 +6^5 y^5 ...
$$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}={x}^{\mathrm{2}} …{i} \\ $$$$\frac{{y}^{\mathrm{2}} +\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}=\mathrm{6}^{\mathrm{5}} {y}^{\mathrm{5}} …{ii} \\ $$$$\frac{{z}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{20}^{\mathrm{7}} {z}^{\mathrm{7}} \\ $$$${i}×{ii}×{iii}: \\ $$$$\mathrm{6}^{\mathrm{5}} .\mathrm{20}^{\mathrm{7}} .{x}^{\mathrm{2}} {y}^{\mathrm{5}} {z}^{\mathrm{7}} =\mathrm{1} \\ $$$${i}\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}={x}^{\mathrm{2}} {y}^{\mathrm{2}} +{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{6}^{\mathrm{5}} .\mathrm{20}^{\mathrm{7}} .\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right){y}^{\mathrm{3}} {z}^{\mathrm{7}} =\mathrm{1} \\ $$$${y}^{\mathrm{3}} {z}^{\mathrm{7}} =\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{5}} .\mathrm{20}^{\mathrm{7}} } \\ $$$${ii}\Rightarrow{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{6}^{\mathrm{5}} {y}^{\mathrm{5}} {z}^{\mathrm{2}} +\mathrm{6}^{\mathrm{5}} {y}^{\mathrm{5}} \\ $$$$… \\ $$
Answered by Rasheed.Sindhi last updated on 30/Jun/25
(√((x^2 +1 )/(y^2 +1))) =x...i 5(√((y^2 +1)/(z^2 +1))) =6y...ii 7(√((z^2 +1)/(x^2 +1))) =20z...iii i×ii×iii: 1×5×7=x×6y×20z xyz=((35)/(120))=(7/(24)) ⇒z=(7/(24xy)) ii⇒ 25(((y^2 +1)/(((7/(24xy)))^2 +1))) =36y^2 ((25(y^2 +1))/(36y^2 ))=((7/(24xy)))^2 +1...v iii⇒ 49(((((7/(24xy)))^2 +1)/(x^2 +1))) =20^2 ((7/(24xy)))^2 ((400×49)/(49×576x^2 y^2 ))=((7/(24xy)))^2 +1...vi From v & vi ((25(y^2 +1))/(36y^2 ))= ((400×49)/(49×576x^2 y^2 ))=((25)/(36x^2 y^2 )) ((y^2 +1)/y^2 )=(1/(x^2 y^2 )) x^2 y^2 +x^2 =1 x^2 =(1/(y^2 +1)) i⇒((x^2 +1 )/(y^2 +1)) =x^2 (((1/(y^2 +1))+1)/(y^2 +1))=(1/(y^2 +1)) (1/(y^2 +1))+1=1 ...
$$\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{1}\:}{{y}^{\mathrm{2}} +\mathrm{1}}}\:={x}…{i} \\ $$$$\:\:\mathrm{5}\sqrt{\frac{{y}^{\mathrm{2}} +\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}}\:=\mathrm{6}{y}…{ii} \\ $$$$\:\mathrm{7}\sqrt{\frac{{z}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\:=\mathrm{20}{z}…{iii} \\ $$$${i}×{ii}×{iii}: \\ $$$$\mathrm{1}×\mathrm{5}×\mathrm{7}={x}×\mathrm{6}{y}×\mathrm{20}{z} \\ $$$${xyz}=\frac{\mathrm{35}}{\mathrm{120}}=\frac{\mathrm{7}}{\mathrm{24}}\:\:\:\Rightarrow{z}=\frac{\mathrm{7}}{\mathrm{24}{xy}} \\ $$$${ii}\Rightarrow\:\mathrm{25}\left(\frac{{y}^{\mathrm{2}} +\mathrm{1}}{\left(\frac{\mathrm{7}}{\mathrm{24}{xy}}\right)^{\mathrm{2}} +\mathrm{1}}\right)\:=\mathrm{36}{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\frac{\mathrm{25}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{36}{y}^{\mathrm{2}} }=\left(\frac{\mathrm{7}}{\mathrm{24}{xy}}\right)^{\mathrm{2}} +\mathrm{1}…{v} \\ $$$${iii}\Rightarrow\:\mathrm{49}\left(\frac{\left(\frac{\mathrm{7}}{\mathrm{24}{xy}}\right)^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:=\mathrm{20}^{\mathrm{2}} \left(\frac{\mathrm{7}}{\mathrm{24}{xy}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{400}×\mathrm{49}}{\mathrm{49}×\mathrm{576}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }=\left(\frac{\mathrm{7}}{\mathrm{24}{xy}}\right)^{\mathrm{2}} +\mathrm{1}…{vi} \\ $$$${From}\:{v}\:\&\:{vi} \\ $$$$\:\frac{\mathrm{25}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{36}{y}^{\mathrm{2}} }=\:\frac{\mathrm{400}×\mathrm{49}}{\mathrm{49}×\mathrm{576}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{36}{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\frac{{y}^{\mathrm{2}} +\mathrm{1}}{{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$${i}\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{1}\:}{{y}^{\mathrm{2}} +\mathrm{1}}\:={x}^{\mathrm{2}} \\ $$$$\:\:\:\:\frac{\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}=\mathrm{1} \\ $$$$\:\:\:\:\: \\ $$$$… \\ $$
Answered by mr W last updated on 30/Jun/25
x, y, z>0 (i)×(ii)×(ii): 5×7=6×20xyz ⇒xyz=(7/(24)) (i)^2 : x^2 +1=x^2 +x^2 y^2 1=x^2 y^2 ⇒xy=1 ⇒z=(7/(24)) (iii): x=(√((z^2 +1)((7/(20z)))^2 −1)) =(√(((7^2 /(24^2 ))+1)(((7×24)/(20×7)))^2 −1))=(3/4) ⇒y=(1/x)=(4/3)
$${x},\:{y},\:{z}>\mathrm{0} \\ $$$$\left({i}\right)×\left({ii}\right)×\left({ii}\right): \\ $$$$\mathrm{5}×\mathrm{7}=\mathrm{6}×\mathrm{20}{xyz} \\ $$$$\Rightarrow{xyz}=\frac{\mathrm{7}}{\mathrm{24}} \\ $$$$\left({i}\right)^{\mathrm{2}} : \\ $$$${x}^{\mathrm{2}} +\mathrm{1}={x}^{\mathrm{2}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\mathrm{1}={x}^{\mathrm{2}} {y}^{\mathrm{2}} \:\Rightarrow{xy}=\mathrm{1} \\ $$$$\Rightarrow{z}=\frac{\mathrm{7}}{\mathrm{24}} \\ $$$$\left({iii}\right): \\ $$$${x}=\sqrt{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{20}{z}}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:=\sqrt{\left(\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{24}^{\mathrm{2}} }+\mathrm{1}\right)\left(\frac{\mathrm{7}×\mathrm{24}}{\mathrm{20}×\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{{x}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Answered by Ghisom last updated on 01/Jul/25
because all (√(...))>0 we can square and let x^2 =u∧y^2 =v∧z^2 =w then it′s straight forward with no tricks needed... (1) ((u+1)/(v+1))=u (2) ((25(v+1))/(w+1))=36v (3) ((49(w+1))/(u+1))=400w ⇒ (1) v=(1/u) (2) w=((25u−11)/(36)) (3) u=(9/(16)) ⇒v=((16)/9)∧w=((49)/(576)) x=(3/4)∧y=(4/3)∧z=(7/(24))
$$\mathrm{because}\:\mathrm{all}\:\sqrt{…}>\mathrm{0}\:\mathrm{we}\:\mathrm{can}\:\mathrm{square}\:\mathrm{and}\:\mathrm{let} \\ $$$${x}^{\mathrm{2}} ={u}\wedge{y}^{\mathrm{2}} ={v}\wedge{z}^{\mathrm{2}} ={w}\:\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{straight} \\ $$$$\mathrm{forward}\:\mathrm{with}\:\mathrm{no}\:\mathrm{tricks}\:\mathrm{needed}… \\ $$$$\left(\mathrm{1}\right)\:\frac{{u}+\mathrm{1}}{{v}+\mathrm{1}}={u} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{25}\left({v}+\mathrm{1}\right)}{{w}+\mathrm{1}}=\mathrm{36}{v} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{49}\left({w}+\mathrm{1}\right)}{{u}+\mathrm{1}}=\mathrm{400}{w} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:{v}=\frac{\mathrm{1}}{{u}} \\ $$$$\left(\mathrm{2}\right)\:{w}=\frac{\mathrm{25}{u}−\mathrm{11}}{\mathrm{36}} \\ $$$$\left(\mathrm{3}\right)\:{u}=\frac{\mathrm{9}}{\mathrm{16}}\:\Rightarrow{v}=\frac{\mathrm{16}}{\mathrm{9}}\wedge{w}=\frac{\mathrm{49}}{\mathrm{576}} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{4}}\wedge{y}=\frac{\mathrm{4}}{\mathrm{3}}\wedge{z}=\frac{\mathrm{7}}{\mathrm{24}} \\ $$

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