Question

(√2+1) square
can you do it
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Answer 1

Answer:

3 + 2√2

Step-by-step explanation:

${( \sqrt{2} + 1)}^{2} \\ \\ = {( \sqrt{2} )}^{2} + {(1)}^{2} + 2 \times \sqrt{2} \times 1 \\ \\ as \: \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ = 2 + 1 + 2 \sqrt{2} \\ \\ = 3 + 2 \sqrt{2}$

Another method is in the attached image.

Answer 2

$\huge{\bold{\red{\underline{\overline{\mid{Solution}\mid}}}}}$

We know that,

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

So,

$= > {( \sqrt{2} + 1) }^{2} \\ = > {( \sqrt{2} )}^{2} + {(1)}^{2} + 2 \times \sqrt{2} \times 1 \\ = > 2 + 1 + 2 \sqrt{2}$

=> 3 + 2√2

Some more identities:-

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ {(a}^{2} - {b}^{2} ) = (a + b)(a - b) \\ \\ {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) \\ \\ {(a - b)}^{3} = {a}^{ 3} - {b}^{3} - 3ab(a - b)$

Hope it helps u...