Question

pls answer to this question i will mark brainliest...pls frnds​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

let us assume that $\sqrt{3}+\sqrt{7}$ is a rational number.

then,

$\sqrt{3} + \sqrt{7} = \frac{p}{q}$

$\sqrt{3} = \frac{p}{q} - \sqrt{7}$

SQUARING ON BOTH SIDES,

${( \sqrt{3} )}^{2} = { (\frac{p}{q} - \sqrt{7} ) }^{2}$

$3 = \frac{ {p}^{2} }{ {q}^{2} } + 7 - \frac{2p \sqrt{7} }{q}$

$\frac{2p \sqrt{7} }{q} = \frac{ {p}^{2} }{ {q}^{2} } + 7 - 3$

$\frac{2p \sqrt{7} }{q} = \frac{ {p}^{2} }{ {q}^{2} } + 4$

$\frac{2p \sqrt{7} }{q} = \frac{ {p}^{2} + 4 {q}^{2} }{ {q}^{2} }$

$\sqrt{7} = \frac{ {p}^{2} + 4 {q}^{2} }{ {q}^{2} } \times \frac{q}{2p}$

$\sqrt{7} = \frac{ {p}^{2} + 4 {q}^{2} }{ 2pq }$

$\frac{ {p}^{2} + 4 {q}^{2} }{ 2pq }$ is rational so $\sqrt{7}$ will also be rational.But this contradicts the fact that $\sqrt{7}$is irrational.This contradiction has arisen due to our wrong assumption.

HENCE,$\sqrt{3}+\sqrt{7}$ is an irrational number.

HOPE THAT HELPS!!