Question

Show that:
$(\frac{1}{2} )^{4} + (\frac{1}{3}) ^{4} + (\frac{1}{6})^{2} \div (\frac{1}{2})^{2} + (\frac{1}{ 3})^{2} + (\frac{1}{6}) - \frac{1}{36} = \frac{1}{6}$
​

Answer 1

Correct Question :-

check weather (1/2)⁴ + (1/3)⁴ + (1/6)² ÷ (1/2)² + (1/3)² + (1/6) - (1/36) = (1/6) or not ?

Solution :-

Solving LHS Part ,

→ (1/2)⁴ + (1/3)⁴ + (1/6)² ÷ (1/2)² + (1/3)² + (1/6) - (1/36)

According to BODMAS RULE Lets First Solve Divide Part ,

→ (1/2)⁴ + (1/3)⁴ + (1/6)² ÷ (1/2)² + (1/3)² + (1/6) - (1/36)

→ (1/2)⁴ + (1/3)⁴ + [ (1/36) ÷ (1/4) ] + (1/3)² + (1/6) - (1/36)

→ (1/2)⁴ + (1/3)⁴ + [ (1/36) × (4/1) ] + (1/3)² + (1/6) - (1/36)

→ (1/2)⁴ + (1/3)⁴ + (1/9) + (1/3)² + (1/6) - (1/36)

→ (1/16) + (1/81) + (1/9) + (1/9) + (1/6) - (1/36)

→ (1/16) + [ (1/81) + (1/9) + (1/9) ] + [ (1/6) - (1/36) ]

Taking LCM ,

→ (1/16) + [ (1 + 9 + 9) / 81 ] + [ (6 - 1) / 36 ]

→ (1/16) + (19/81) + (5/36)

Taking LCM again Now,

→ (81 + 304 + 180) /1296

→ (565 / 1296 )

→ ≠ RHS.

So, we can conclude That, LHS is Not Equal to RHS.

Answer 2

$\rule{248}{3}$

CORRECT QUESTION :-

$\sf{Check\ whether\ }\displaystyle{\sf{\frac{(\frac{1}{2} )^4+(\frac{1}{3} )^4+(\frac{1}{6} )^2}{(\frac{1}{2} )^2+(\frac{1}{3} )^2+(\frac{1}{6} )^2}-\frac{1}{36} =\displaystyle{\sf{\frac{1}{6}. }}}}$

$\rule{248}{3}$

SOLUTION :-

LHS :-

$\displaystyle{\sf{\frac{(\frac{1}{2} )^4+(\frac{1}{3} )^4+(\frac{1}{6} )^2}{(\frac{1}{2} )^2+(\frac{1}{3} )^2+(\frac{1}{6} )^2}-\frac{1}{36} }}$

RHS :-

$\displaystyle{\sf{\frac{1}{6} }}$

Let us solve the LHS.

$\displaystyle{\sf{=\frac{\frac{1}{16} +\frac{1}{81} +\frac{1}{36} }{\frac{1}{4} +\frac{1}{9} +\frac{1}{36} }-\frac{1}{36} }}\\\\\\\displaystyle{\sf{=\frac{\frac{81+16+36}{1296} }{\frac{9+4+1}{36} } -\frac{1}{36} }}\\\\\\\displaystyle{\sf{=\frac{\frac{133}{1296} }{\frac{14}{36} }-\frac{1}{36} }}\\\\\\\displaystyle{\sf{=({\frac{133}{1296} \times \frac{36}{14} )-\frac{1}{36} }}$

$\displaystyle{\sf{=(\frac{133}{36}\times \frac{1}{14} ) -\frac{1}{36} }}\\\\\\\displaystyle{\sf{=\frac{133}{504} -\frac{1}{36} }}\\\\\\\displaystyle{\sf{=\frac{133-14}{504} }}\\\\\\\displaystyle{\sf{=\frac{119}{504} }}\\\\\\\displaystyle{\sf{=\frac{7}{72} }}$

∴ So, we conclude that, LHS ≠ RHS.

$\rule{248}{3}$