Question

if lmn=2, then prove that (1/2+l+2m^-1)+(2^-1/1+m+n^-1)+(2^-1/1+n+2l^-1)=1/2​

Answer 1

Correct Question (1) :-

if lmn = 2, then prove that (1/2+l+2m^-1)+(2^-1/1+m+n^-1)+2^(-1)/(1+n+2l^-1) = 1/2

Solution :-

→ (1/2+l+2m^-1) + (2^-1/1+m+n^-1) + (2^-1/1+n+2l^-1) = 1/2

Solving Each part of LHS seperately,

First Part :- 1/{2+l+2m^(-1)}

→ 1/{2+l+2m^(-1)}

→ 1/{2+l+(2/m)}

→ 1/{(2m + lm + 2) / m}

→ m/(lm + 2m + 2)

Second Part :- 2^(-1) / {1+m+n^(-1)}

→ 2^(-1) / {1+m+n^(-1)}

→ 1/2{1+m+n^(-1)}

→ 1/2(1 + m + 1/n)

→ 1/(2 + 2m + 2/n)

Now, we have given that, lmn = 2

So, lm = (2/n)

Putting value now,

→ 1/(2 + 2m + lm)

→ 1/(lm + 2m + 2)

Third Part :- (2)^(-1)/{1 + n + 2l^(-1)}

→ (1/2)/{1 + n + (2/l)}

Multiply & Divide by lm in numerator and denominator now,

→ (lm/2) / {lm + lmn + 2m}

Putting lmn = 2 in denominator now,

→ (lm/2) / (lm + 2m + 2)

Adding all three parts now,

→ m/(lm + 2m + 2) + 1/(lm + 2m + 2) + (lm/2) / (lm + 2m + 2)

As denominator is same ,

→ {m + 1 + (lm/2)} / (lm + 2m + 2)

→ {(2m + 2 + lm) /2} / (lm + 2m + 2)

→ (1/2) * [ (lm + 2m + 2) / (lm + 2m + 2) ]

→ (1/2) * 1

→ (1/2) (Ans.)

___________________

Correct Question (2) :- if lmn = 1 , then prove that, 1/{1 + l + m^(-1)} + 1/{1 + m + n^(-1)} + 1/{1 + n + l^(-1)} = 1 ?

Solution :-

→ 1/{1 + l + m^(-1)} + 1/{1 + m + n^(-1)} + 1/{1 + n + l^(-1)} = 1

Solving Each part of LHS seperately,

First Part :- 1/{1 + l + m^(-1)}

→ 1/{1 + l + m^(-1)}

Multiply numerator and denominator by m ,

→ m * 1 / m * {1 + l + m^(-1)}

→ m / (m + lm + 1)

→ m/(lm + m + 1)

Second Part :- 1/{1 + m + n^(-1)}

→ 1/{1 + m + n^(-1)}

→ 1/{1 + m + (1/n)}

Now, we have given that, lmn = 1

So, lm = (1/n)

Putting value now,

→ 1/(1 + m + lm)

→ 1/(lm + m + 1)

Third Part :- 1/{1 + n + l^(-1)}

→ 1/{1 + n + l^(-1)}

Again, using lmn = 1 ,

So, n = (1/lm)

putting value now,

→ 1/{ 1 + (1/lm) + (1/l)}

→ 1/{(lm + 1 + m) / lm}

→ lm/(lm + m + 1)

Now, Putting all three parts we get :-

→ 1/{1 + l + m^(-1)} + 1/{1 + m + n^(-1)} + 1/{1 + n + l^(-1)}

→ {m/(lm + m + 1)} + {1/(lm + m + 1)} + {lm/(lm + m + 1)}

Taking LCM now,

→ {(m + 1 + lm) /(lm + m + 1)}

→ { (lm + m + 1)} / (lm + m + 1) }

→ 1 = RHS . (Proved).

____________________

Answer 2

Given : lmn=2,

To Find : prove that (1/2+l+2m^-1)+(2^-1/1+m+n^-1)+(2^-1/1+n+2l^-1)=1/2

Solution:

lmn=2

1/(2+l+2m⁻¹)

=(1/(2 + l + 2/m)

= m/(2m + lm + 2)

2⁻¹/(1+m+n⁻¹)

= 1/(2 + 2m + 2/n)

2/n = lm

= 1/(2 + 2m +lm)

2⁻¹/(1+n+2l⁻¹)

1/(2  + 2n  + 4/l)

= (1/2)/(1 + n + 2/l)

= (lm/2)/(lm + lmn + 2m)

= (lm/2)/(lm + 2 + 2m)

Adding all

(m + 1 + (lm/2) ) / ( (lm + 2 + 2m)

= ((2m + 2 + lm)/2 ) /  ( (lm + 2 + 2m)

= 1/2

= RHS

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