Question

Answer for this question

Answer 1

Proofs provided below

Explanation:

`\bold {\text{Prove } (a + b)^2 = a^2 + 2ab + b^2}}`

`\\\\\implies (a + b)^2 = (a+b) * (a+b)\\\\\implies (a + b)^2 = a * (a+b) +b * (a+b)\\\\\implies (a + b)^2 = a * a + a * b + b * a + b * b\\\\\implies (a + b)^2 = a^2+ab+ba+b^2\\\\\implies (a + b)^2 = a^2+ab+ab+b^2 \;\;\;\;\text{ since ab = ba}\\\\\implies (a+b)^2 = a^2+2ab+b^2\\\\`

`\bold{\text{Prove } a^2-b^2 \,=\, (a+b)(a-b)\\\\}`

1. Add and subtract ab to LHS

`\implies a^2-b^2 = a^2-b^2-ab+ab\\\\\implies a^2-b^2 = a^2-ab+ab-b^2\\\\`

2. Factorize the above expression

`\implies a^2-b^2 = a(a-b)+b(a-b)\\\\\implies a^2-b^2 = (a-b)(a+b) \;\;\;\; \text{since (a-b) is a common factor in RHS }`

∴ (a² - b²) = (a - b) (a + b)

`\bold{\text{Prove $(√(3)+1)/(√(3)-1)$= $2+\sqrt{\ensuremath{3}}$}}\\`

`\text{1. Multiply LHS by $\frac{\sqrt{\text{}3}-1}{√(3)-1}$}\\\\\implies $\frac{\sqrt{\text{}3}+1}{√(3)-1}$ * $\frac{\sqrt{\text{}3}-1}{√(3)-1}$ \\\\`

`\implies ( (√(3) + 1)(√(3)-1) )/((√(3) - 1)(√(3)-1) )\\\\`

Numerator is

`(√(3) + 1)(√(3)-1) = (√(3))^2 - 1^2 = 3 - 1 = 2\\\\`

Denominator is

`(√(3)-1)^2 = (√(3))^2 - 2\cdot √(3) \;\cdot 1 + (-1)^2\\\\= 3 - 2 √(3) + 1\\\\= 4 - 2 √(3)\\\\`

So LHS becomes

`(2)/(4 - 2√(3)) \\\\`

Dividing numerator and denominator by 2 yields

`(2)/(4 - 2√(3)) = \frac {1}{2 - √(3)}`

Multiply numerator and denominator by
`{2-\sqrt {3}}`

`$(1\cdot(2+√(3)))/((2-√(3))(2+√(3)))$`

`=$\frac{\ensuremath{2}+√(3)}{2^(2)-(\sqrt{3)^(2)}}=$$\frac{\ensuremath{\ensuremath{2}+√(3)}}{4-3}=\ensuremath{2}+√(3)$`

Hence Proved