Question

prove that one upon root 4 + root 5 + 1 upon root 5 + root 6 + 1 upon root 6 + root 7 + 1 upon root 7 + root 8 + 1 upon root 8 + 9 equal 1​

Answer 1

Answer:

Cannot equal.

Remember

Square Root ≥ 0

Since root 4 = 2

Their sum ≥ 2 > 1.

Therefore, it cannot equal.

Answer 2

$\sf\huge\bold{Answer:}$

Given :

$\tt\ \frac{ \: \: \: 1}{ \sqrt{4} } + \sqrt{5} + \frac{1}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \sqrt{8} + \frac{1}{ \sqrt{8} } = 1$

To prove :

• LHS=RHS or not

Solution :

• LHS

$\tt\ = \frac{ \: \: \: 1}{ \sqrt{4} } + \sqrt{5} + \frac{1}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \sqrt{8} + \frac{1}{ \sqrt{8} }$

• Taking LCM

$\tt\ = \frac{ \: \: \: 1}{ \sqrt{4} } + \frac{ (\sqrt{5} \times \sqrt{5} )+ 1}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \frac{ (\sqrt{8} \times \sqrt{8} )+ 1}{ \sqrt{8} }$

$\tt\ = \frac{ \: \: \: 1}{ \sqrt{4} } + \frac{ \sqrt{5 \times 5} + 1}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \frac{ \sqrt{8 \times 8} + 1}{ \sqrt{8} }$

$\tt\ = \frac{ \: \: \: 1}{ \sqrt{4} } + \frac{ 5 + 1}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \frac{ 8 + 1}{ \sqrt{8} }$

$\tt\ = \frac{ \: \: \: 1}{ \sqrt{4} } + \frac{ \: \: \: 6}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \frac{ \: \: \: 9}{ \sqrt{8} }$

$\tt\ = \frac{ \: \: \: 1}{ \sqrt{4} } + \frac{ \: \: \: 6}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \frac{ \sqrt{7} }{1} + \frac{ \: \: \: 9}{ \sqrt{8} }$

• Simplifying by remove square roots

$\tt\ = \frac{1}{ 2 } + \frac{ \: \: \: 6}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \sqrt{7} + \frac{ \: \: \: 9}{ 2\sqrt{2} }$

$\tt\ = \frac{1}{ 2 } + \frac{ \: \: \: 9}{ 2\sqrt{2} } + \frac{ \: \: \: 6}{ \sqrt{5} } + \frac{1}{ \sqrt{6} } + \frac{ \sqrt{7} }{1}$

• Taking LCM

$\tt\ = \bigg( \frac{ \sqrt{2} + 9}{ 2\sqrt{2} } \bigg) + \bigg(\frac{ 6 \sqrt{6} + 1 \sqrt{5} }{ \sqrt{5} \times \sqrt{6}} \bigg) + \frac{ \sqrt{7} }{1}$

$\tt\ = \bigg( \frac{ \sqrt{2} + 9}{ 2\sqrt{2} } \bigg) + \bigg(\frac{ 6 \sqrt{6} + \sqrt{5} }{ \sqrt{30} } \bigg) + \frac{ \sqrt{7} }{1}$

• Rationalising denominator

$\tt\ = \bigg( \frac{ \sqrt{2} + 9}{ 2\sqrt{2} } \times \frac{2 \sqrt{2} }{ 2\sqrt{2} } \bigg) + \bigg(\frac{ 6 \sqrt{6} + \sqrt{5} }{ \sqrt{30} } \times \frac{ \sqrt{30} }{ \sqrt{30} } \bigg) + \frac{ \sqrt{7} }{1}$

$\tt\ = \bigg( \frac{2 \times \sqrt{2} \times \sqrt{2} + 9 }{2 \times 2 \times \sqrt{2} \times \sqrt{2} } \bigg) + \bigg(\frac{6 \sqrt{6} \times \sqrt{30} + \sqrt{5} \times \sqrt{30} }{ \sqrt{30} \times \sqrt{30} } \bigg) + \frac{ \sqrt{7} }{1}$

$\tt\ = \bigg( \frac{2 \times 2 + 9 }{4 \times 2 } \bigg) + \bigg(\frac{6 \sqrt{6} \times \sqrt{5 \times 6} + \sqrt{5} \times \sqrt{5 \times 6} }{ 30 } \bigg) + \frac{ \sqrt{7} }{1}$

$\tt\ = \bigg( \frac{4 + 9 }{8 } \bigg) + \bigg(\frac{6 \times {6} \times \sqrt{5 } + {5} \times \sqrt{ 6} }{ 30 } \bigg) + \frac{ \sqrt{7} }{1}$

$\tt\ = \bigg( \frac{13 }{8 } \bigg) + \bigg(\frac{36 \sqrt{5 } + {5} \sqrt{ 6} }{ 30 } \bigg) + \frac{ \sqrt{7} }{1}$

• Taking LCM

$\tt\ = \frac{30 \times 13 + 8(36 \sqrt{5} + 5 \sqrt{6} ) + \sqrt{7} \times 8 \times 30}{8 \times 30}$

$\tt\ = \frac{390 + 288 \sqrt{5} + 40 \sqrt{6} + 240 \sqrt{7} }{240}$

• Putting values of √5,√6 & √7

$\tt\ = \frac{390 + 288 \times 2.23 + 40 \times 2.44 + 240 \times 2.64 }{240}$

$\tt\ = \frac{390 + 643.9 + 97.9 + 634.9 }{240}$

$\tt\ = \frac{1766.7}{240}$

$\tt\ =7.36125$

$\tt\ ≈7$

• RHS

$\tt\ = 1$

$\rm \red {LHS≠RHS}$

Hence, the above statement is incorrect.