4-m-1-4-2-2-m-2-2-m-1-m-

Question Number 222001 by fantastic last updated on 14/Jun/25

$$\left(\frac{\mathrm{4}^{{m}+\frac{\mathrm{1}}{\mathrm{4}}} ×\sqrt{\mathrm{2}.\mathrm{2}^{{m}} }}{\mathrm{2}.\sqrt{\mathrm{2}^{−{m}} }}\right)^{\frac{\mathrm{1}}{{m}}} =?? \\ $$

Commented by fantastic last updated on 16/Jun/25

$${Oooo} \\ $$

Commented by MATHEMATICSAM last updated on 15/Jun/25

$$\mathrm{You}\:\mathrm{read}\:\mathrm{in}\:\mathrm{class}\:\mathrm{9}\:\mathrm{wb}\:\mathrm{board}? \\ $$

Commented by fantastic last updated on 15/Jun/25

$${how}\:{did}\:{you}… \\ $$

Commented by MATHEMATICSAM last updated on 15/Jun/25

because these maths were in our wb board text book

$$\mathrm{because}\:\mathrm{these}\:\mathrm{maths}\:\mathrm{were}\:\mathrm{in}\:\mathrm{our}\:\mathrm{wb}\:\mathrm{board} \\ $$$$\mathrm{text}\:\mathrm{book} \\ $$

Commented by fantastic last updated on 16/Jun/25

$${Are}\:{you}\:{in}\:\mathrm{10}^{{th}} {or}\:\mathrm{11}^{{th}} ?? \\ $$

Commented by MATHEMATICSAM last updated on 16/Jun/25

$$\mathrm{11th} \\ $$

Answered by Rasheed.Sindhi last updated on 14/Jun/25

(((4^(m+(1/4)) ×(√(2.2^m )))/(2.(√2^(−m) ))))^(1/m) =?? =((((2^2 )^((4m+1)/4) ×2^((m+1)/2) )/(2.2^(−(m/2)) )))^(1/m) =(((2^((4m+1)/2) ×2^((m+1)/2) )/2^(−(m/2)+1) ))^(1/m) =((2^((5m+2)/2) /2^((−m+2)/2) ))^(1/m) =(2^(((5m+2)/2)−((−m+2)/2)) )^(1/m) =(2^((6m)/2) )^(1/m) =2^(3m×(1/m)) =2^3 =8

$$\left(\frac{\mathrm{4}^{{m}+\frac{\mathrm{1}}{\mathrm{4}}} ×\sqrt{\mathrm{2}.\mathrm{2}^{{m}} }}{\mathrm{2}.\sqrt{\mathrm{2}^{−{m}} }}\right)^{\frac{\mathrm{1}}{{m}}} =?? \\ $$$$=\left(\frac{\left(\mathrm{2}^{\mathrm{2}} \right)^{\frac{\mathrm{4}{m}+\mathrm{1}}{\mathrm{4}}} ×\mathrm{2}^{\frac{{m}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}.\mathrm{2}^{−\frac{{m}}{\mathrm{2}}} }\right)^{\frac{\mathrm{1}}{{m}}} \\ $$$$=\left(\frac{\mathrm{2}^{\frac{\mathrm{4}{m}+\mathrm{1}}{\mathrm{2}}} ×\mathrm{2}^{\frac{{m}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}^{−\frac{{m}}{\mathrm{2}}+\mathrm{1}} }\right)^{\frac{\mathrm{1}}{{m}}} \\ $$$$=\left(\frac{\mathrm{2}^{\frac{\mathrm{5}{m}+\mathrm{2}}{\mathrm{2}}} }{\mathrm{2}^{\frac{−{m}+\mathrm{2}}{\mathrm{2}}} }\right)^{\frac{\mathrm{1}}{{m}}} \\ $$$$=\left(\mathrm{2}^{\frac{\mathrm{5}{m}+\mathrm{2}}{\mathrm{2}}−\frac{−{m}+\mathrm{2}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{{m}}} \\ $$$$=\left(\mathrm{2}^{\frac{\mathrm{6}{m}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{{m}}} \\ $$$$=\mathrm{2}^{\mathrm{3}{m}×\frac{\mathrm{1}}{{m}}} \\ $$$$=\mathrm{2}^{\mathrm{3}} \\ $$$$=\mathrm{8} \\ $$

Commented by fantastic last updated on 15/Jun/25

$$\mathbb{N}\boldsymbol{\mathrm{i}}\mathbb{C}\epsilon \\ $$

Commented by Rasheed.Sindhi last updated on 15/Jun/25

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}! \\ $$

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