Question

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NO 3 AND 4​

Answer 1

Question-1:

Show that: $\sf {(\dfrac {4m}{3}- \dfrac {3n}{4})^{2} + 2mn = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

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Solution:

$\sf {(\dfrac {4m}{3}- \dfrac {3n}{4})^{2} + 2mn = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

$\bigstar {\sf {\orange {Using\ the\ identity, (a-b)^{2} = a^{2} - 2ab + b^{2}}}}$

$\sf {(\dfrac {4m}{3}) ^{2} - 2 \times \dfrac{4m}{3} \times \dfrac{3n}{4} + (\dfrac {3n}{4})^{2} + 2mn = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

$\sf {(\dfrac {4m}{3}) ^{2} - 2 \times \dfrac{\cancel{4}m}{\cancel{3}} \times \dfrac{\cancel{3}n}{\cancel{4}} + (\dfrac {3n}{4})^{2} + 2mn = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

$\sf {(\dfrac {4m}{3}) ^{2} - 2 mn + (\dfrac {3n}{4})^{2} + 2mn = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

$\sf {(\dfrac {4m}{3}) ^{2} \cancel{- 2 mn} + (\dfrac {3n}{4})^{2} \cancel{ + 2 mn} = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

$\sf {(\dfrac {4m}{3}) ^{2} + (\dfrac {3n}{4})^{2} = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

$\sf {\dfrac {16}{9} m ^{2} + \dfrac {9 }{16} n ^{2} = \dfrac {16}{9} m^{2} + \dfrac {9}{16}n^{2}}$

LHS = RHS

Hence, proved!

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Question-2:

Divide: 52pqr (p+q) (q+r) (r+p) ÷ 104pq (q+r) (r+p)

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Solution:

$\sf \dfrac{52pqr (p+q) (q+r) (r+p)}{ 104pq (q+r) (r+p)}$

$\sf \dfrac{ \cancel{52}\cancel{pq}r (p+q) (q+r) (r+p)}{ \cancel{104}\cancel{pq} (q+r) (r+p)}$

$\sf \dfrac{ r (p+q) \cancel{(q+r) }\cancel{(r+p)}}{ 2 \cancel{(q+r) }\cancel{(r+p)}}$

$\implies \sf \dfrac{ r (p+q) }{ 2 }$