hey please help simplify this sum-- it's urgent!!!!
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the required answer is 1/2bc
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Hi ,
i ) 1/a + 1/( b + c ) = ( b + c + a )/ a( b + c )--( 1 )
ii ) 1/a - 1/( b + c ) = ( b + c - a )/a( b + c ) ---( 2 )
iii ) [ 1 + ( b² + c² - a² )/2bc ]
= [ ( 2bc + b² + c² - a² )/2bc ]
= [ ( b + c )² - a² ]/2bc
= [ ( b + c + a )( b + c - a ) ]/2bc ]
= [ ( a + b + c )(b + c - a ) ]/2bc ---( 3 )
iv ) ( a + b + c )^-2
= 1/( a + b + c )² ----- ( 4 )
Now ,
[1/a+1/b+c]/[1/a-1/b+c][1+(b²+c²-a²)/2bc](a+b+c)^-2
[ from ( 1 ), ( 2 ) , ( 3 ) and ( 4 ) ]
= [ ( 1 )/ ( 2 ) ] [ ( 3 ) ] ( 4 )
= [(a+b+c)/(b+c-a)][(a+b+c)(b+c-a)/2bc][1/(a+b+c)²]
= 1/2bc
I hope this helps you.
: )
i ) 1/a + 1/( b + c ) = ( b + c + a )/ a( b + c )--( 1 )
ii ) 1/a - 1/( b + c ) = ( b + c - a )/a( b + c ) ---( 2 )
iii ) [ 1 + ( b² + c² - a² )/2bc ]
= [ ( 2bc + b² + c² - a² )/2bc ]
= [ ( b + c )² - a² ]/2bc
= [ ( b + c + a )( b + c - a ) ]/2bc ]
= [ ( a + b + c )(b + c - a ) ]/2bc ---( 3 )
iv ) ( a + b + c )^-2
= 1/( a + b + c )² ----- ( 4 )
Now ,
[1/a+1/b+c]/[1/a-1/b+c][1+(b²+c²-a²)/2bc](a+b+c)^-2
[ from ( 1 ), ( 2 ) , ( 3 ) and ( 4 ) ]
= [ ( 1 )/ ( 2 ) ] [ ( 3 ) ] ( 4 )
= [(a+b+c)/(b+c-a)][(a+b+c)(b+c-a)/2bc][1/(a+b+c)²]
= 1/2bc
I hope this helps you.
: )
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