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Answer for this question-example-1
User Yanga
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Answer:

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

User Ranuka
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