Question

(2m+1)³+(n+3)³/(2m+1)³-(n+3)=536/193 and n²-m²/n²+m²=155/493 find m and n​

Answer 1

Step-by-step explanation:

$\frac{ {n}^{2} - {m}^{2} }{ {n}^{2} + {m}^{2} } = \frac{155}{493}$

Componendo & Dividendo

$\frac{ {n}^{2} - {m}^{2} - ( {n}^{2} + {m}^{2} ) }{ {n}^{2} - {m}^{2} + ( {n}^{2} + {m}^{2} ) } = \frac{155 - 493}{155 + 493}$

$= > \frac{ - 2 {m}^{2} }{2 {n}^{2} } = \frac{ - 338}{648}$

$= > \frac{ {m}^{2} }{ {n}^{2} } = \frac{338}{648} = \frac{169}{324} = \frac{ {13}^{2} }{ {18}^{2} }$

$= > {( \frac{m}{n} )}^{2} = ({ \frac{13}{18} })^{2}$

$= > \frac{m}{n} = \frac{13}{18}$

$\frac{( {2m + 1})^{3} + ({n + 3})^{3} }{ ({2m + 1})^{3} - ({n + 3})^{3} } = \frac{536}{193}$

Componendo & Dividendo,

$= > \frac{ 2({n + 3})^{3} }{2( {2m + 1})^{3} } = \frac{536 - 193}{536 + 193} = \frac{343}{729}$

$= > ({ \frac{n + 3}{2m + 1} })^{3} = \frac{ {7}^{3} }{ {9}^{3} } = ( { \frac{7}{9} })^{3}$

$= > \frac{n + 3}{2m + 1} = \frac{7}{9}$

$= > 9(n + 3) = 7(2m + 1)$

$= > 9n + 27 = 14m + 7$

$= > 9n = 14m - 20 \: \: \: - > eq1$

We also know that,

$\frac{m}{n} = \frac{13}{18}$

$= > m = ( \frac{13}{18} )n \: \: \: - > eq2$

Substituting Eq 2 in Eq 1, we get

$= > 9n = 14( \frac{13}{18} )n - 20$

$= > 9n = \frac{91n}{9} - 20$

$= > \frac{91n}{9} - 9n = 20$

$= > \frac{91n - 81n}{9} = 20$

$= > \frac{10n}{9} = 20$

$= > \frac{n}{9} = 2$

$= > n = 18$

$= > m = ( \frac{13}{18} )n = \frac{13}{18} \times 18 = 13$

Therefore,

• m = 13
• n = 18

Hope this helps!