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Question Number 96505 by bemath last updated on 02/Jun/20

((√((8)^(1/(4 )) −(√((√2)+1))))/((√((8)^(1/(4 )) +(√((√2)−1))))−(√((8)^(1/(4 )) −(√((√2)−1)))))) ?

$$\frac{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}}{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}+\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}−\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}}\:? \\ $$

Commented by bemath last updated on 02/Jun/20

yes sir. you are right

$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$

Commented by bobhans last updated on 02/Jun/20

maybe it ((√((8)^(1/(4 )) −(√((√2)+1))))/((√((8)^(1/(4 )) +(√((√2)−1))))−(√((8)^(1/(4 )) −(√((√2)−1))))))

$$\mathrm{maybe}\:\mathrm{it}\:\frac{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}}{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}+\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}−\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}} \\ $$

Answered by bobhans last updated on 02/Jun/20

w = ((√((8)^(1/(4 )) −(√((√2)+1))))/((√((8)^(1/(4 )) +(√((√2)−1))))−(√((8)^(1/(4 )) −(√((√2)−1)))))) w^2 = (((8)^(1/(4 )) −(√((√2)−1)))/(2 (8)^(1/(4 )) −2(√(((64))^(1/(4 )) −((√2)−1))))) w^2 = (((8)^(1/(4 )) −(√((√2)−1)))/(2((8)^(1/(4 )) −(√((√2)−1))))) = (1/2) w = (1/(√2))

$$\mathrm{w}\:=\:\frac{\sqrt{\sqrt[{\mathrm{4}\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}}{\sqrt{\sqrt[{\mathrm{4}\:}]{\mathrm{8}}+\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}−\sqrt{\sqrt[{\mathrm{4}\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}} \\ $$$$\mathrm{w}^{\mathrm{2}} \:=\:\frac{\sqrt[{\mathrm{4}\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\mathrm{2}\:\sqrt[{\mathrm{4}\:}]{\mathrm{8}}−\mathrm{2}\sqrt{\sqrt[{\mathrm{4}\:}]{\mathrm{64}}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}} \\ $$$$\mathrm{w}^{\mathrm{2}} \:=\:\frac{\sqrt[{\mathrm{4}\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\mathrm{2}\left(\sqrt[{\mathrm{4}\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{w}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\: \\ $$

Answered by 1549442205 last updated on 02/Jun/20

Putting^4 (√8)=(√(2(√2)))=a,(√((√2)−1))=b we have a^2 −b^2 =(√2)+1.Hence, A=((√(a−(√(a^2 −b^2 ))))/((√(a+b))−(√(a−b))))=((√(a−(√(a^2 −b^2 ))))/((√(((√(a+b))−(√(a−b)))))^2 )) =((√(a−(√(a^2 −b^2 ))))/(√(2a−2(√(a^2 −b^2 )))))=((√(a−(√(a^2 −b^2 ))))/((√2)((√(a−(√(a^2 −b^2 )))))) =(1/(√2)).Thus,A=(1/(√2))

$$\mathrm{Putting}\:^{\mathrm{4}} \sqrt{\mathrm{8}}=\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{a},\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{b} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} =\sqrt{\mathrm{2}}+\mathrm{1}.\mathrm{Hence}, \\ $$$$\mathrm{A}=\frac{\sqrt{\mathrm{a}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}}{\sqrt{\mathrm{a}+\mathrm{b}}−\sqrt{\mathrm{a}−\mathrm{b}}}=\frac{\sqrt{\mathrm{a}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}}{\left.\sqrt{\left(\sqrt{\mathrm{a}+\mathrm{b}}−\sqrt{\mathrm{a}−\mathrm{b}}\right.}\right)^{\mathrm{2}} } \\ $$$$=\frac{\sqrt{\mathrm{a}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}}{\sqrt{\mathrm{2a}−\mathrm{2}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}}=\frac{\sqrt{\mathrm{a}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}}{\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{a}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}\right.} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}.\mathrm{Thus},\mathrm{A}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$

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